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$$ \int \frac{1}{\sin{x}\cos^{3}x}dx $$

$$\Rightarrow \int \frac{1}{\sin{x}\cos^{3}x}{\cos{x}\over \cos{x}}dx$$

$$\Rightarrow \int \frac{\sec^{4}{x}}{\tan{x}}dx$$

$$\Rightarrow \int \frac{\sec^{2}(1+\tan^{2}x)}{\tan{x}}dx$$

$$Substitution \tan{x}=t \Rightarrow \sec^{2}xdx=dt$$

$$\Rightarrow \int \frac{1+t^{2}}{t}dt$$

$$\Rightarrow \int \frac{1}{t}dt+\int tdt $$

$$\Rightarrow \log|t|+\frac{t^{2}}{2}+C$$

$$\Rightarrow \log|\tan{x}|+\frac{1}{2}\tan^{2}x+C$$

Any other elegant way to do this?

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    That looks pretty elegant to me! You could also do $\int \frac{\sin x}{(1-\cos^2x)\cos^3x} \,dx = \int \frac{du}{(u^2-1)u^3}$ and then use partial fractions; but I think this takes longer. – 2017-02-10

2 Answers 2

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Another way, I am not sure it is an elegant way but is different than yours $$\int \frac { 1 }{ \sin { x } \cos ^{ 3 } x } dx=\int { \frac { \sin ^{ 2 }{ x+\cos ^{ 2 }{ x } } }{ \sin { x } \cos ^{ 3 } x } dx= } \int { \frac { \sin { x } }{ \cos ^{ 3 }{ x } } } dx+\int { \frac { 1 }{ \sin { x\cos { x } } } dx } =\\ =-\int { \frac { d\cos { x } }{ \cos ^{ 3 }{ x } } +\int { \frac { \sin ^{ 2 }{ x+\cos ^{ 2 }{ x } } }{ \sin { x } \cos { x } } } } dx=\frac { 1 }{ 2\cos ^{ 2 }{ x } } +\int { \frac { \sin { x } }{ \cos { x } } dx+\int { \frac { \cos { x } }{ \sin { x } } dx } = } \\ =\frac { 1 }{ 2\cos ^{ 2 }{ x } } -\ln { \left| \cos { x } \right| +\ln { \left| \sin { x } \right| } +C= } \frac { 1 }{ 2\cos ^{ 2 }{ x } } +\ln { \left| \tan { x } \right| } +C$$

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    i didn't get the first integral of second line. What's $ d\cos{x} $? What you did there? – 2017-02-10
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    @mikealise, instead of making substitution as $$\\ \sin { xdx } =du\quad ,-\cos { x } =u\\ \int { \frac { \sin { x } }{ \cos ^{ 3 }{ x } } } dx=-\int { \frac { du }{ { u }^{ 3 } } } $$ you can do this directly as i did.$$\int { \frac { \sin { x } }{ \cos ^{ 3 }{ x } } } dx=-\int { \frac { d\left( \cos { x } \right) }{ \cos ^{ 3 }{ x } } } $$ – 2017-02-10
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    how did $\int \frac{d\cos{x}}{cos^{3}x}$ simplify to $\frac{1}{2cos^{2}x}$ – 2017-02-10
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    $$\sin { xdx } =du\quad ,-\cos { x } =u\\ \int { \frac { \sin { x } }{ \cos ^{ 3 }{ x } } } dx=-\int { \frac { du }{ { u }^{ 3 } } } =???\\ $$ – 2017-02-10
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    cheers mate! that was helpful – 2017-02-10
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    you're welcome) – 2017-02-10
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This is not better than your solution but works: \begin{eqnarray} \int \frac{1}{\sin{x}\cos^{3}x}dx &=& \int \frac{\cos x}{\sin{x}\cos^{4}x}dx &=& \int \frac{du}{u(1-u^2)^2}dx \end{eqnarray}