Mathematics

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


SOLUTION
Solution :-

$$\displaystyle I= \int \frac{dx}{sinx\, cos^{3}x}$$
$$\displaystyle I= \int sin^{-1}x.cos^{-3}xdx$$

Divide Num & Denominator by $$cos^{4}x$$
$$\displaystyle \therefore I= \int \dfrac{1/cos^{4}x.dx}{\dfrac{sinx.cos^{3}x}{cos^{4}x}}$$
$$\displaystyle =\int \frac{sec^{4}xdx}{tan x}$$

Let $$tan x= t$$

$$\Rightarrow sec^{2}xdx= dt$$
$$\displaystyle \therefore I= \int \frac{sec^{2}x.dt}{t}$$
$$\displaystyle =\int \frac{(1+tan^{2}x).dt}{t}$$
$$\displaystyle =\int \frac{(1+t^{2})dt}{t}$$

$$\displaystyle =\int (\frac{1}{t}+t)dt$$
$$\displaystyle = logt+\frac{t^{2}}{2}+c$$

$$\displaystyle = log\left | tanx \right |+\frac{tan^{2}x}{2}+c$$
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Subjective Medium Published on 17th 09, 2020
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