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Integration by Parts Worked Exercises


{$ \color{blue}{y= xe^{x^2} } $}
{$ \color{red}{y= {1 \over 2 } e^{x^2} +0} $}
{$ \color{green}{y= {1 \over 2 } e^{x^2} +1} $}
{$ \color{brown}{y= {1 \over 2 } e^{x^2} +2} $}
{$ \color{lime}{y= {1 \over 2 } e^{x^2} -1} $}
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1) {$ \large \int xe^{x^2} dx $}

For integration by parts,

{$ \large \int uv' dx = uv - \int u'v dx, $} the immediate observation is that the integrand is almost the derivative of {$ e^{x^2}$}, which would be {$ 2xe^{x^2} $}. Choosing {$ u={ 1 \over 2 }$} solves it neatly, since the derivative of u is zero which causes the integral part on the right to disappear.

u{$ 1 \over 2 $}
u'{$ 0 $}
v{$e^{x^2} $}
v'{$2xe^{x^2} $}

{$$ \large \eqalign { \int uv' dx &= uv - \int u'v dx \\ \int \frac{1}{2} 2 x e^{x^2} dx &= \frac{1}{2} e^{x^2} - \int 0\ e^{x^2} dx \\ \int x e^{x^2} dx &= \frac{1}{2} e^{x^2} - \int 0 dx \\ \therefore \int x e^{x^2} dx &= \frac{1}{2} e^{x^2} + C }$$}

Check: {$ {d \over {dx}}\left[ \frac{1}{2} e^{x^2} + C \right] = \frac{1}{2} (2x e^{x^2}) + 0 = xe^{x^2}$}


{$ \color{blue}{y= t\sin t }$}
{$\color{red}{y= -t \cos t + sin t +0} $}
{$ \color{green}{y= -t \cos t + sin t +1} $}
{$ \color{brown}{y= -t \cos t + sin t +2} $}
{$ \color{lime} {y= -t \cos t + sin t -1} $}
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2) {$ \large \int t\ sin\ t\ dt $}

u{$ -t $}
u'{$ -1 $}
v{$ -\cos t $}
v'{$ \sin t $}


{$$ \large \eqalign{ \int uv' dt &= uv - \int u'v dt \\ \int -t \sin t dt &= -t (- \cos t) - \int -1 (-\cos t) dt \\ -\int t \sin t dt &= t\cos t - \int \cos t dt \\ \therefore \int t \sin t dt &= -t\cos t + \sin t + C }$$}

Check:

{$$ \eqalign { \frac{d}{dt}\left[ -t\cos t + \sin t + C \right] &= \\ \frac{d}{dt}\left[ -t\cos t\right] + \frac{d}{dt}\left[\sin t\right] + \frac{d}{dt}\left[C \right] &= \\ -t\frac{\cos t}{dt} + \frac{-t}{dt}\cos t + \cos t + 0 &= \\ -t(-\sin t) - \cos t + \cos t + 0 &= t\sin t } $$}


{$ \color{blue}{y= x^2e^{2x} }$}
{$ \color{red}{y= \frac{x^2e^{2x}}{2} -\frac{xe^{2x}}{2} + \frac{e^{2x}}{4} +0} $}
{$ \color{green}{y= \frac{x^2e^{2x}}{2} -\frac{xe^{2x}}{2} + \frac{e^{2x}}{4} +1 } $}
{$ \color{brown}{y= \frac{x^2e^{2x}}{2} -\frac{xe^{2x}}{2} + \frac{e^{2x}}{4} +2} $}
{$ \color{lime}{y= \frac{x^2e^{2x}}{2} -\frac{xe^{2x}}{2} + \frac{e^{2x}}{4} -1} $}
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3) {$ \large \int x^2e^{2x} dx $}

u{$ x^2 $}
u'{$ 2x $}
v{$ \frac{e^{2x}}{2} $}
v'{$ e^{2x} $}
w{$ x $}
w'{$ 1 $}

{$$ \large \eqalign{ \int uv' dx &= uv - \int u'v dx \\ \int x^2 e^{2x} dx &= \frac{x^2e^{2x}}{2} - \int xe^{2x} dx }$$}

At this point it may be necessary to integrate the rightmost integral by parts.

{$$ \large \eqalign{ \int wv' dx &= wv - \int w'v dx \\ \int xe^{2x} dx &= \frac{xe^{2x}}{2} - \int \frac{e^{2x}}{2} dx \\ \int xe^{2x} dx &= \frac{xe^{2x}}{2} - \frac{e^{2x}}{4} + C }$$}

This result may be substituted in the previous one.

{$$ \large \eqalign{ \int x^2 e^{2x} dx &= \frac{x^2e^{2x}}{2} - \int xe^{2x} dx \\ \int x^2 e^{2x} dx &= \frac{x^2e^{2x}}{2} - (\frac{xe^{2x}}{2} - \frac{e^{2x}}{4} ) +C \\ \therefore \int x^2 e^{2x} &= \frac{x^2e^{2x}}{2} - \frac{xe^{2x}}{2} + \frac{e^{2x}}{4} +C }$$}

Check: {$$ \eqalign { \frac{d}{dx}\left[ \frac{x^2e^{2x}}{2} - \frac{xe^{2x}}{2} + \frac{e^{2x}}{4} +C \right] &= \\ \frac{d}{dx}\left[ \frac{x^2e^{2x}}{2} \right] - \frac{d}{dx}\left[\frac{xe^{2x}}{2}\right] + \frac{d}{dx}\left[\frac{e^{2x}}{4}\right] +\frac{d}{dx}\left[C \right] &= \\ \frac{1}{2}\left[ x^2\frac{d}{dx}e^{2x} + e^{2x}\frac{d}{dx}x^2 \right] - \frac{1}{2}\left[x\frac{d}{dx}e^{2x}+e^{2x}\frac{x}{dx}\right] + 2\frac{e^{2x}}{4} +0 &= \\ \frac{1}{2}\left[ x^2 2e^{2x} + e^{2x}2x \right] - \left[xe^{2x}+\frac{e^{2x}}{2}\right] + \frac{e^{2x}}{2} &= \\ x^2e^{2x} + xe^{2x} - xe^{2x}-\frac{e^{2x}}{2} + \frac{e^{2x}}{2} &= x^2e^{2x} }$$}


{$ \color{blue}{y= \arcsin x }$}
{$ \color{red}{y= \arcsin\ x + \sqrt{1 - x^2} +0} $}
{$ \color{green}{y= \arcsin\ x + \sqrt{1 - x^2} +1 } $}
{$ \color{brown}{y= \arcsin\ x + \sqrt{1 - x^2} +2} $}
{$ \color{lime}{y= \arcsin\ x + \sqrt{1 - x^2} -1} $}
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4) {$ \large \int \arcsin\ x\ dx $}

u{$ \arcsin x $}
u'{$ \frac{1}{\sqrt{1 - x^2}}$}
v{$ x $}
v'{$ 1 $}

{$$ \eqalign{ \int uv' dx &= uv - \int u'v dx \\ \int \arcsin\ x dx &= x\ \arcsin\ x - \int \frac{x\ dx}{\sqrt{1 - x^2}} \\ \int \arcsin\ x dx &= x\ \arcsin\ x - \int x(1 - x^2)^{-\frac{1}{2}} \\ \int \arcsin\ x dx &= x\ \arcsin\ x + (1 - x^2)^{\frac{1}{2}} + C \\ \therefore \int \arcsin\ x dx &= x\ \arcsin\ x + \sqrt{1 - x^2} + C }$$}

Check: {$$ \eqalign { \frac{d}{dx}\left[x\ \arcsin\ x + \sqrt{1 - x^2} + C \right] &= \\ \frac{d}{dx}\left[x\arcsin\ x \right] + \frac{d}{dx}\left[\sqrt{1 - x^2}\right] + \frac{d}{dx}\left[C \right] &= \\ \frac{d}{dx}\left[x\right]\arcsin\ x + x\frac{d}{dx}\left[arcsin\ x \right] + \frac{d}{dx}\left[(1 - x^2)^\frac{1}{2}\right] + 0 &= \\ \arcsin\ x + \frac{x}{\sqrt(1 - x^2)} + (-\frac{1}{2})2x(1 - x^2)^{-\frac{1}{2}} + 0 &= \\ \arcsin\ x + \frac{x}{\sqrt{(1 - x^2)}} -\frac{x}{\sqrt{(1 - x^2)}} &= \arcsin\ x}$$}


{$ \color{blue}{y= x^3\log\ x }$}
{$ \color{red}{y= \frac{x^4}{4}(\log\ x - \frac{1}{4}) +0} $}
{$ \color{green}{y= \frac{x^4}{4}(\log\ x - \frac{1}{4}) +1 } $}
{$ \color{brown}{y= \frac{x^4}{4}(\log\ x - \frac{1}{4}) +2} $}
{$ \color{lime}{y= \frac{x^4}{4}(\log\ x - \frac{1}{4}) -1} $}
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5) {$ \large \int x^3\log\ x \ dx$}

u{$ \log\ x $}
u'{$ \frac{1}{x} $}
v{$ \frac{x^4}{4} $}
v'{$ x^3 $}

{$$ \large \eqalign{ \int uv' dx &= uv - \int u'v dx \\ \int x^3 \log x\ dx &= \frac{x^4}{4}\log\ x - \int \frac{1}{x}\frac{x^4}{4} dx \\ \int x^3 \log x\ dx &= \frac{x^4}{4}\log\ x - \int \frac{x^3}{4} dx \\ \int x^3 \log x\ dx &= \frac{x^4}{4}\log\ x - \frac{x^4}{16} + C\\ \therefore \int x^3 \log x\ dx &= \frac{x^4}{4}(\log\ x - \frac{1}{4}) + C}$$}

Check: {$$ \eqalign { \frac{d}{dx}\left[ \frac{x^4}{4}(\log\ x - \frac{1}{4}) + C \right] &= \\ \frac{d}{dx}\left[ \frac{x^4}{4}\right] (\log\ x - \frac{1}{4}) + \frac{x^4}{4}\frac{d}{dx}\left[ (\log\ x - \frac{1}{4})\right] + \frac{d}{dx}\left[ C \right] &= \\ x^3\log\ x - \frac{x^3}{4} + \frac{x^4}{4} (\frac{1}{x} - 0) +0 &= \\ x^3\log\ x - \frac{x^3}{4} + \frac{x^3}{4} &= x^3\log\ x }$$}


{$ \color{blue}{y= u\sqrt{1-u} }$}
{$\color{red}{y= -\frac{2}{3}u\sqrt{(1-u)^3} - \frac{4}{15}\sqrt{(1-u)^5} +0} $}
{$ \color{green}{y= -\frac{2}{3}u\sqrt{(1-u)^3} - \frac{4}{15}\sqrt{(1-u)^5} + 1 } $}
{$\color{brown}{y= -\frac{2}{3}u\sqrt{(1-u)^3} - \frac{4}{15}\sqrt{(1-u)^5} + 2} $}
{$ \color{lime}{y= -\frac{2}{3}u\sqrt{(1-u)^3} - \frac{4}{15}\sqrt{(1-u)^5} -1} $}
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6) {$ \large \int u\sqrt{1-u}\ dx$}

w{$ u $}
w'{$ 1 $}
v{$ -\frac{2}{3}\sqrt{(1-u)^3} $}
v'{$ \sqrt{1-u} $}


{$$ \large \eqalign{ \int wv' du &= wv - \int w'v du \\ \int u(1-u)^{\frac{1}{2}} du &= -\frac{2}{3}u\sqrt{(1-u)^3} - \int -\frac{2}{3}\sqrt{(1-u)^3}du \\ \int u(1-u)^{\frac{1}{2}} du &= -\frac{2}{3}u\sqrt{(1-u)^3} + \frac{2}{3}(-\frac{2}{5})\sqrt{(1-u)^5} + C \\ \therefore \int u(1-u)^{\frac{1}{2}} du &= -\frac{2}{3}u\sqrt{(1-u)^3} - \frac{4}{15}\sqrt{(1-u)^5} + C \\ }$$}

Check: {$$ \eqalign { \frac{d}{du}\left[ -\frac{2}{3}u\sqrt{(1-u)^3} - \frac{4}{15}\sqrt{(1-u)^5} + C \right] &= \\ \frac{d}{du}\left[ -\frac{2}{3}u\sqrt{(1-u)^3} \right] - \frac{d}{du}\left[ \frac{4}{15}\sqrt{(1-u)^5} \right] + \frac{d}{du}\left[ C \right] &= \\ \frac{d}{du}\left[u\right]( -\frac{2}{3}\sqrt{(1-u)^3}) - u\frac{d}{du}\left[ \frac{2}{3}\sqrt{(1-u)^3} \right] - (- \frac{2}{3}\sqrt{(1-u)^3}) &= \\ -\frac{2}{3}\sqrt{(1-u)^3}+ u\sqrt{1-u} + \frac{2}{3}\sqrt{(1-u)^3}) &= u\sqrt{1-u} }$$}


{$ \color{blue}{y= \int x^2\ \arcsin\ }$}
{$ \color{red}{y= \frac{1}{3}x^3 \arcsin x + \frac{1}{9}x^2\sqrt{1-x^2} + \frac{2}{9}\sqrt{1-x^2} +0} $}
{$ \color{green}{y= \frac{1}{3}x^3 \arcsin x + \frac{1}{9}x^2\sqrt{1-x^2} + \frac{2}{9}\sqrt{1-x^2} + 1 } $}
{$ \color{brown}{y= \frac{1}{3}x^3 \arcsin x + \frac{1}{9}x^2\sqrt{1-x^2} + \frac{2}{9}\sqrt{1-x^2} + 2} $}
{$\color{lime}{y= \frac{1}{3}x^3 \arcsin x + \frac{1}{9}x^2\sqrt{1-x^2} + \frac{2}{9}\sqrt{1-x^2} -1} $}
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7) {$ \large \int x^2\ \arcsin\ x\ dx $}

u{$ x^2 $}
u'{$ 2x $}
v{$ x \arcsin x + \sqrt{1-x^2} $}
v'{$ \arcsin x $}

In this exercise I use a previous result and a little algebra because I could do the larger amounts of algebra that integrating by parts again required.
{$$ \eqalign{ \int uv' dx &= uv - \int 2x dx \\ \int x^2 \arcsin x dx &= x^2\left[x \arcsin x + \sqrt{1-x^2} \right] - \int 2x\left[x \arcsin x + \sqrt{1-x^2} \right]dx \\ \int x^2 \arcsin x dx &= x^3 \arcsin x + x^2\sqrt{1-x^2} - 2\int x^2 \arcsin x - \int 2x\sqrt{1-x^2} dx \\ }$$}

This might seem like the wrong track because the integral we are solving for has appeared on the right. But a little addition and division fixes that.

{$$ \eqalign{ 3\int x^2 \arcsin x dx &= x^3 \arcsin x + x^2\sqrt{1-x^2} - \int 2x\sqrt{1-x^2} dx \\ \int x^2 \arcsin x dx &= \frac{1}{3}x^3 \arcsin x + \frac{1}{3}x^2\sqrt{1-x^2} - \frac{1}{3}\int 2x\sqrt{1-x^2} dx \\ \int x^2 \arcsin x dx &= \frac{1}{3}x^3 \arcsin x + \frac{1}{3}x^2\sqrt{1-x^2} - \frac{1}{3}\left[ -\frac{2}{3}\sqrt{(1-x^2)^3} \right] +C \\ \int x^2 \arcsin x dx &= \frac{1}{3}x^3 \arcsin x + \frac{1}{3}x^2\sqrt{1-x^2} + \frac{2}{9}\left[(1-x^2)\sqrt{1-x^2} \right] +C \\ \int x^2 \arcsin x dx &= \frac{1}{3}x^3 \arcsin x + \left(\frac{1}{3}-\frac{2}{9}\right)x^2\sqrt{1-x^2} + \frac{2}{9}\sqrt{1-x^2} +C \\ \therefore \int x^2 \arcsin x dx &= \frac{1}{3}x^3 \arcsin x + \frac{1}{9}x^2\sqrt{1-x^2} + \frac{2}{9}\sqrt{1-x^2} +C \\ }$$}

Check: {$$ \eqalign { \frac{d}{dx}\left[ \frac{1}{3}x^3 \arcsin x + \frac{1}{9}x^2\sqrt{1-x^2} + \frac{2}{9}\sqrt{1-x^2} +C \right] &= \\ \frac{d}{dx}\left[ \frac{1}{3}x^3 \arcsin x \right]+ \frac{d}{dx}\left[\frac{1}{9}x^2\sqrt{1-x^2}\right] + \frac{d}{dx}\left[\frac{2}{9}\sqrt{1-x^2}\right] +\frac{d}{dx}\left[C \right] &= \\ \frac{d}{dx}\left[ \frac{1}{3}x^3 \arcsin x \right]+ \frac{d}{dx}\left[\frac{1}{9}x^2\sqrt{1-x^2}\right] -\frac{2x}{9\sqrt{1-x^2}}+ 0 &= \\ \frac{d}{dx}\left[ \frac{1}{3}x^3 \arcsin x \right]+ \frac{d}{dx}\left[\frac{1}{9}x^2\right]\sqrt{1-x^2}+ \frac{1}{9}x^2\frac{d}{dx}\left[\sqrt{1-x^2}\right] -\frac{2x}{9\sqrt{1-x^2}} &= \\ \frac{d}{dx}\left[ \frac{1}{3}x^3 \arcsin x \right]+ \frac{d}{dx}\left[\frac{1}{9}x^2\right]\sqrt{1-x^2}- \frac{x2}{9\sqrt{1-x^2}} -\frac{2x}{9\sqrt{1-x^2}} &= \\ \frac{d}{dx}\left[ \frac{1}{3}x^3 \arcsin x \right]+ \frac{2x}{9}\sqrt{1-x^2}- \frac{xx^2}{9\sqrt{1-x^2}} -\frac{2x}{9\sqrt{1-x^2}} &= \\ \frac{d}{dx}\left[ \frac{1}{3}x^3\right]\arcsin x + \frac{1}{3}x^3\frac{d}{dx}\left[\arcsin x \right]+ \frac{2x(1-x^2)}{9\sqrt{1-x^2}}- \frac{x^3}{9\sqrt{1-x^2}} -\frac{2x}{9\sqrt{1-x^2}} &= \\ x^2\arcsin x + \frac{x^3}{3\sqrt{1-x^2}}+ \frac{2x(1-x^2)}{9\sqrt{1-x^2}}- \frac{x^3}{9\sqrt{1-x^2}} -\frac{2x}{9\sqrt{1-x^2}} &= \\ x^2\arcsin x + \frac{3x^3}{9\sqrt{1-x^2}}+ \frac{2x-2x^3}{9\sqrt{1-x^2}}- \frac{x^3}{9\sqrt{1-x^2}} -\frac{2x}{9\sqrt{1-x^2}} &= \\ x^2\arcsin }$$}


{$ \color{blue}{y= x\ \tan^2x\ }$}
{$ \color{red}{y= x \tan x - \frac{x^2}{2} - \log \cos x +0} $}
{$ \color{green}{y= x \tan x - \frac{x^2}{2} - \log \cos x + 1 } $}
{$ \color{brown}{y= x \tan x - \frac{x^2}{2} - \log \cos x + 2} $}
{$ \color{lime}{y= x \tan x - \frac{x^2}{2} - \log \cos x-1} $}
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8) {$ \large \int x\ \tan^2x\ dx $}

u{$ x $}
u'{$ 1 $}
v{$ \tan x - x $}
v'{$ \tan^2 x $}


{$$ \large \eqalign{ \int uv' dx &= uv - \int u'v dx \\ \int \tan^2 dx &= x(\tan x - x) - \int \tan x - x dx \\ \int \tan^2 dx &= x \tan x - x^2 - \int \tan x d + \int x dx \\ \int \tan^2 dx &= x \tan x - x^2 - \int \tan x d + \int x dx \\ \int \tan^2 dx &= x \tan x - x^2 - \log \cos x + \frac{x^2}{2} + C \\ \therefore \int \tan^2 dx &= x \tan x - \frac{x^2}{2} - \log \cos x + C \\ }$$}

Check: {$$ \eqalign { \frac{d}{dx}\left[x \tan x - \frac{x^2}{2} - \log \cos x + C \right] &= \\ \frac{d}{dx}\left[x \tan x\right] - \frac{d}{dx}\left[\frac{x^2}{2}\right] - \frac{d}{dx}\left[\log \cos x \right] + \frac{d}{dx}\left[C \right] &= \\ \frac{d}{dx}\left[x \right] \tan x + x \frac{d}{dx}\left[\tan x\right] - x + \frac{-\sin x}{\cos x} + 0 &= \\ \tan x + x \sec^2x - x - \tan x &= \\ x (\sec^2x - 1) &= x\tan^2 x }$$}


{$ \color{blue}{y= x^3\sqrt{a^2-x^2} }$}

for several values of a. Other colors:
solutions of the integral
for various choices of a and C.
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9) {$ \large \int x^3\sqrt{a^2-x^2}\ dx$}

u{$ x^2 $}
u'{$ 2x $}
v{$ -\frac{1}{3} \sqrt{(a^2-x^2)^3}$}
v'{$x \sqrt{a^2-x^2} $}

The reason for splitting the {$ x^3 $} this way is that avoiding an {$ x $} factor in v' would be very difficult.
{$$ \large \eqalign{ \int uv' dx &= uv - \int u'v dx \\ \int x^3\sqrt{a^2-x^2}\ dx &= -\frac{x^2}{3} \sqrt{(a^2-x^2)^3} - \int -\frac{2x}{3} \sqrt{(a^2-x^2)^3} dx \\ \int x^3\sqrt{a^2-x^2}\ dx &= -\frac{x^2}{3} \sqrt{(a^2-x^2)^3} - \frac{2x}{3}(-2x)\frac{2}{5} \sqrt{(a^2-x^2)^5} + C\\ \therefore \int x^3\sqrt{a^2-x^2}\ dx &= -\frac{x^2}{3} \sqrt{(a^2-x^2)^3} - \frac{2}{15} \sqrt{(a^2-x^2)^5} + C \\ }$$}

Check: {$$ \eqalign { \frac{d}{dx}\left[-\frac{x^2}{3} \sqrt{(a^2-x^2)^3} - \frac{2}{15} \sqrt{(a^2-x^2)^5} + C\right]&= \\ \frac{d}{dx}\left[ -\frac{x^2}{3} \sqrt{(a^2-x^2)^3}\right] - \frac{d}{dx}\left[ \frac{2}{15} \sqrt{(a^2-x^2)^5} \right] + \frac{d}{dx}\left[ C\right] &= \\ -\frac{2x}{3}\sqrt{(a^2-x^2)^3} -\frac{x^2}{3}\frac{d}{dx}\left[\sqrt{(a^2-x^2)^3}\right] - \frac{2}{15}(-2x)\frac{5}{2} \sqrt{(a^2-x^2)^5} &= \\ -\frac{2x}{3}\sqrt{(a^2-x^2)^3} -\frac{x^2}{3}\frac{3}{2}(-2x)\sqrt{a^2-x^2} + \frac{2x}{3}\sqrt{(a^2-x^2)^3} &= x^3\sqrt{a^2-x^2 } }$$}


{$ \color{blue}{y= x^3\sqrt{a^4-x^4} }$}

for several values of a. Other colors:
solutions of the integral
for various choices of a and C.
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10) {$ \large \int x^3\sqrt{a^4-x^4}\ dx$}

u{$ \frac{1}{4} $}
u'{$ 0 $}
v{$ -\frac{2}{3}\sqrt{(a^4-x^4)^3} $}
v'{$ 4x^3\sqrt{a^4-x^4} $}

Choosing u so that the integral on the right disappears seems like cheating, but so far as I can tell, the result checks out.


{$$ \large \eqalign{ \int uv' dx &= uv - \int u'v dx \\ \int \frac{1}{4}4x^3\sqrt{a^4-x^4} dx &= -\frac{1}{4} \frac{2}{3}\sqrt{(a^4-x^4)^3} - \int -0\frac{2}{3}\sqrt{(a^4-x^4)^3} dx \\ \therefore \int x^3\sqrt{a^4-x^4} dx &= -\frac{1}{6}\sqrt{(a^4-x^4)^3} + C}$$}

Check: {$$ \eqalign { \frac{d}{dx}\left[-\frac{1}{6}\sqrt{(a^4-x^4)^3} + C \right] &= \\ -\frac{1}{6}\frac{3}{2}(-4x^3)\sqrt{a^4-x^4} + \frac{d}{dx}\left[C \right] &= 4x^3\sqrt{a^4-x^4} }$$}


{$ \color{blue}{y= \cos\theta\log\sin\theta } $}
{$ \color{red}{y= sin\theta(\log\sin\theta -1) +0} $}
{$ \color{green}{y= sin\theta(\log\sin\theta -1) +1} $}
{$ \color{brown}{y= sin\theta(\log\sin\theta -1) +2} $}
{$ \color{lime}{y= sin\theta(\log\sin\theta -1) -1}$}

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11) {$ \large \int \cos \theta\ \log\sin\theta\ d\theta$}

u{$ \log \sin \theta $}
u'{$ \frac{\cos\theta}{\sin\theta} $}
v{$ \sin\theta $}
v'{$ \cos\theta $}


This is just a reminder: {$ \log $} here always means the natural logarithm unless another base is explicitly given.

{$$ \large \eqalign{ \int uv' d\theta &= uv - \int d\theta \\ \int \cos\theta\log \sin \theta d\theta &= \sin\theta \log \sin \theta - \int \sin\theta\frac{\cos\theta}{\sin\theta} d\theta \\ \int \cos\theta\log \sin \theta d\theta &= \sin\theta \log \sin \theta - \int \cos\theta d\theta \\ \int \cos\theta\log \sin \theta d\theta &= \sin\theta \log \sin \theta - \sin\theta + C \\ \therefore \int \cos\theta\log \sin \theta d\theta &= \sin\theta( \log \sin \theta - 1) + C \\}$$}

Check: {$$ \eqalign { \frac{d}{dx}\left[\sin\theta( \log \sin \theta - 1) + C \right] &= \\ \frac{d}{dx}\left[\sin\theta\right]( \log \sin \theta - 1) + \sin\theta\frac{d}{dx}\left[( \log \sin \theta - 1)\right] + \frac{d}{dx}\left[C \right] &= \\ \cos\theta \log \sin \theta - \cos\theta + \sin\theta\ \frac{\cos\theta}{\sin\theta} + 0 &= \cos\theta \log \sin \theta }$$}


{$ \color{blue}{y= \sec^4\theta } $}
{$ \color{red}{y= \frac{1}{3}\tan^3\theta + \tan\theta }$}
{$ \color{green}{y= \frac{1}{3}\tan^3\theta + \tan\theta +1} $}
{$ \color{brown}{y= \frac{1}{3}\tan^3\theta + \tan\theta +2} $}
{$ \color{lime}{y= \frac{1}{3}\tan^3\theta + \tan\theta -1}$}
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12) {$ \large \int \sec^4\theta\ d\theta$}

{$$ \large \eqalign{ \int \sec^4\theta\ d\theta &= \int (\tan^2\theta +1)\sec^2\theta d\theta \\ &= \int \tan^2\theta\sec^2\theta d\theta + \int \sec^2\theta d\theta \\ &= \int \tan^2\theta\sec^2\theta d\theta + \tan\theta +C \\}$$}

Now to solve the remaining integral by parts.

u{$ \tan^2\theta $}
u'{$ 2 \tan\theta\sec^2\theta$}
v{$ \tan\theta$}
v'{$ \sec^2\theta $}

{$$ \large \eqalign { \int uv' d\theta &= uv - \int u'v d\theta \\ \int \tan^2\theta\sec^2\theta d\theta &= \tan^2\theta]tan\theta - 2\int \tan^2\theta\sec^2\theta d\theta \\ 3\int \tan^2\theta\sec^2\theta d\theta &= \tan^3\theta \\ \int \tan^2\theta\sec^2\theta d\theta &= \frac{1}{3}\tan^3\theta \\ }$$}

And substituting in the previous result:

{$$ \large \therefore \int \sec^4\theta\ d\theta = \frac{1}{3}\tan^3\theta + \tan\theta +C $$}

Check: {$$ \eqalign { \frac{d}{dx}\left[ \frac{1}{3}\tan^3\theta + \tan\theta +C \right] &= \\ \frac{d}{dx}\left[ \frac{1}{3}\tan^3\theta \right] + \frac{d}{dx}\left[\tan\theta \right] + \frac{d}{dx}\left[ C \right] &= \\ \frac{1}{3} 3sec^2\theta\tan^2\theta + \sec^2\theta &= \\ sec^2\theta(\tan^2\theta + 1) &= \sec^4 \\ }$$}


{$ \color{blue}{y= \cos^3\theta\ } $}
{$ \color{red}{y= \cos\theta - \frac{1}{3}sin^3\theta +0} $}
{$ \color{green}{y= \cos\theta - \frac{1}{3}sin^3\theta +1} $}
{$ \color{brown}{y= \cos\theta - \frac{1}{3}sin^3\theta +2} $}
{$ \color{lime}{y= \cos\theta - \frac{1}{3}sin^3\theta -1}$}
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13) {$ \large \int \cos^3\theta\ d\theta$}

{$$ \large \eqalign { \int \cos^3\theta\ d\theta &= \\ \int \cos\theta\ (\cos^3\theta\ ) d\theta &= \\ \int \cos\theta\ (1- \sin^2\theta\ ) d\theta &= \\ \int \cos\theta\ - \int \cos\theta\sin^2\theta\ d\theta &= \\ \sin\theta\ - \int \cos\theta\sin^2\theta\ d\theta }$$}

u{$ \sin^2\theta $}
u'{$ 2\sin\theta\cos\theta $}
v{$ \sin\theta $}
v'{$ \cos\theta $}

The remaining integral is worked by parts.

{$$ \large \eqalign { \int uv' d\theta &= uv - \int u'v d\theta \\ \int \cos\theta\sin^2\theta d\theta &= \sin\theta\sin^2\theta - \int 2\cos\theta\sin^2\theta d\theta \\ 3\int \cos\theta\sin^2\theta d\theta &= \sin^3\theta \\ \int \cos\theta\sin^2\theta d\theta &= \frac{1}{3}\sin^3\theta + C\\}$$}

{$$ \large \eqalign { \therefore \int \cos^3\theta\ d\theta &= \sin\theta\ - \frac{1}{3}\sin^3\theta }$$}

Check: {$$ \eqalign { \frac{d}{dx}\left[ \sin\theta\ - \frac{1}{3}\sin^3\theta + C \right] &= \\ \cos -\frac{1}{3}(3)\cos\theta\sin^2\theta &= \\ \cos(1 -\sin^2\theta &= \cos^3\theta }$$}


{$ \color{blue}{y= \frac{x^2}{(a^2-x^2)^{3 \over 2}} } $}

for several values of a. Other colors:
solutions of the integral
for various choices of a and C.
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14) {$ \large \int \frac{x^2dx}{(a^2-x^2)^{3 \over 2} } $}

u{$ x $}
u'{$ 1 $}
v{$ \frac{1}{\sqrt{a^2-x^3}}$}
v'{$ \frac{x}{\sqrt{(a^2-x^2)^3}} $}


{$$ \large \eqalign { \int uv' dx &= uv - \int u'v dx \\ \int x \frac{x}{\sqrt{(a^2-x^2)^3}} dx &= x\frac{1}{\sqrt{a^2-x^2}} - \int \frac{1}{\sqrt{a^2-x^2}} dx \\ \int x \frac{x}{\sqrt{(a^2-x^2)^3}} dx &= \frac{x}{\sqrt{a^2-x^2}} - \frac{x}{\sqrt{(a^2-x^2)^3}} + C \\ }$$}

Check: {$$ \eqalign { \frac{d}{dx}\left[ \frac{x}{\sqrt{a^2-x^2}} - \frac{x}{\sqrt{(a^2-x^2)^3}} + C \right] &= \\ (a^2 -x^2)^{-\frac{1}{2}} + x(x(a^2-x^2)^{-\frac{3}{2}}) - (a^2-x^2)^{-\frac{1}{2}} &= \\ \frac{x^2}{(a^2-x^2)^{3 \over 2}} }$$}


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