Mathematics

$$\displaystyle \int \dfrac{\sec^2 x}{\sec x+\tan x)^5}dx=$$


SOLUTION
$$\displaystyle \int \frac{\sec^{2}x\ dx}{(\sec x+\tan x)^{5}}$$
Take $$\sec x$$ out in denominator
$$\displaystyle =\int \frac{\sec^{2}x \ dx}{\sec^{5}x (1+ \sin x)^{5}}$$

$$\displaystyle =\int \frac{\cos^{3}x \ dx}{(1+\sin x)^{5}}$$
Put $$1+ \sin x=z$$
$$\cos x \ dx = dz$$
$$\displaystyle =\int \frac{1-z^{2}}{z^{5}}dz$$
$$\displaystyle =\int \frac{1}{z^{5}}-\frac{1}{z^{3}}dz$$
$$\displaystyle =-\frac{1}{4{z^{4}}}+\frac{1}{2{z^{2}}}+C$$
$$\displaystyle =-\frac{1}{4(\sin x+1)^{4}}+\frac{1}{2(\sin x+1)^{2}}+C$$
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Subjective Medium Published on 17th 09, 2020
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