Q114 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
$\int\left(\frac{\left(\sin ^4 x+2 \cos ^2 x-1\right) \cos x}{(1+\sin x)^6}\right) d x=$
- A$\frac{\sin ^6 x}{6(1+\sin x)^6}+c$
- B$-\frac{\sin ^6 x}{6(1+\sin x)^6}+c$
- C$\frac{\cos ^6 x}{6(1+\sin x)^6}+c$
- Correct answerD$-\frac{\cos ^6 x}{6(1+\sin x)^6}+c$
Solution
$\begin{aligned}
& \text { } I=\int \frac{\left(\sin ^4 x+2 \cos ^2 x-1\right)}{(1+\sin x)^6} \cos x d x \\
& =\int \frac{\left(\sin ^4 x-2 \sin ^2 x+1\right)}{(1+\sin x)^6} \cos x d x \\
& =\int \frac{\left(1-\sin ^2 x\right)^2}{(1+\sin x)^6} \cos x d x=\int \frac{(1-\sin x)^2}{(1+\sin x)^4} \cos x d x
\end{aligned}$
Let $\sin x=t \Rightarrow \cos x d x=d t$ $\Rightarrow I=\int \frac{(1-t)^2}{(1+t)^4} d t$
Let $t=\cos 2 p \Rightarrow d t=-2 \sin 2 p d p$ $\begin{aligned} & \Rightarrow I=\int \frac{\left(2 \sin ^2 p\right)^2}{\left(2 \cos ^2 p\right)^4}(-2 \sin 2 p) d p \\ & =-\int \frac{4}{16} \frac{\sin ^4 p}{\cos ^8 p}(2 \times 2 \sin p \cos p) d p=-\int \frac{\sin ^5 p}{\cos ^7 p} d p \\ & I=-\int \tan ^5 p \sec ^2 p d p \\ & \Rightarrow I=\frac{-\tan ^6 p}{6}+C \Rightarrow I=\frac{-1}{6}\left(\frac{1-t}{1+t}\right)^3+C \\ & {\left[\because \cos 2 p=t \Rightarrow \frac{1-\tan ^2 p}{1+\tan ^2 p}=t \Rightarrow \tan ^2 p=\frac{1-t}{1+t}\right]} \\ & \Rightarrow I=\frac{-1}{6}\left(\frac{1-\sin x}{1+\sin x}\right)^3+C \Rightarrow I=\frac{-1}{6} \frac{\cos ^6 x}{(1+\sin x)^6}+C . \end{aligned}$
Let $\sin x=t \Rightarrow \cos x d x=d t$ $\Rightarrow I=\int \frac{(1-t)^2}{(1+t)^4} d t$
Let $t=\cos 2 p \Rightarrow d t=-2 \sin 2 p d p$ $\begin{aligned} & \Rightarrow I=\int \frac{\left(2 \sin ^2 p\right)^2}{\left(2 \cos ^2 p\right)^4}(-2 \sin 2 p) d p \\ & =-\int \frac{4}{16} \frac{\sin ^4 p}{\cos ^8 p}(2 \times 2 \sin p \cos p) d p=-\int \frac{\sin ^5 p}{\cos ^7 p} d p \\ & I=-\int \tan ^5 p \sec ^2 p d p \\ & \Rightarrow I=\frac{-\tan ^6 p}{6}+C \Rightarrow I=\frac{-1}{6}\left(\frac{1-t}{1+t}\right)^3+C \\ & {\left[\because \cos 2 p=t \Rightarrow \frac{1-\tan ^2 p}{1+\tan ^2 p}=t \Rightarrow \tan ^2 p=\frac{1-t}{1+t}\right]} \\ & \Rightarrow I=\frac{-1}{6}\left(\frac{1-\sin x}{1+\sin x}\right)^3+C \Rightarrow I=\frac{-1}{6} \frac{\cos ^6 x}{(1+\sin x)^6}+C . \end{aligned}$
Hence, the correct answer is option: 'D'.