Mathematics

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


SOLUTION
$$\displaystyle\int \dfrac{1}{\sin x\cos^3x}dx$$
$$=\displaystyle\int \dfrac{1/\cos^4x}{\dfrac{\sin x\cos^3x}{\cos^4x}}dx$$
$$=\displaystyle\int\dfrac{(1+\tan^2x)\sec^2x}{\tan x}dx$$
Now $$\tan x=t\rightarrow \sec^3xdx=dt$$
$$\Rightarrow \displaystyle\int \dfrac{1+t^2}{t}=dt$$
$$\Rightarrow \displaystyle\int \dfrac{1}{t}i+dt$$
$$\Rightarrow log t+\dfrac{t^2}{2}+c$$
$$\Rightarrow log \tan t+\dfrac{\tan^2x}{2}+c$$.
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Subjective Medium Published on 17th 09, 2020
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