Question

If $I=\int \frac{dx}{(2\sin x+\sec x{)}^4}$, then I equals

• A. $-\frac{1}{5{\tan }^{5}x}+\frac{1}{3{\tan }^{6}x}-\frac{2}{7{\tan }^{7}x}+C$
• B. $\frac{1}{5{\tan }^{5}x}+\frac{1}{3{\tan }^{6}x}-\frac{I}{(2\sin x+\sec x{)}^3}+C$
• C. $\frac{-1}{3(2\sin x+\sec {)}^3}+{\tan }^{-1}\left(3\sqrt{\tan x}\right)+C$
• D. $\frac{-1}{3(2\sin x+\sec x{)}^3}-{\tan }^{-1}\left(3\sqrt{\tan x}\right)+C$

Answer 1

The correct option is A $-\frac{1}{5{\tan }^{5}x}+\frac{1}{3{\tan }^{6}x}-\frac{2}{7{\tan }^{7}x}+C$

$I=\int \frac{dx}{{(2\sin x+\sec x)}^4}=\int \frac{{\sec }^{4}x}{(2\tan x+1+{\tan }^{2}x)}dx$
Put $\tan x+1=t$
$I=\int \frac{{(t-1)}^2+1}{t^8}dt=-\frac{1}{5t^5}+\frac{1}{3t^6}-\frac{2}{7t^7}+c$

$=-\frac{1}{5{\tan }^{5}x}+\frac{1}{3{\tan }^{6}x}-\frac{2}{7{\tan }^{7}x}+C$