MHT CET · Maths · Indefinite Integration
\(\int \cos ^3 x e^{\log (\sin x)^2} d x=\)
- A \(\frac{\sin ^3 x}{3}-\sin ^5 x+c\)
- B \(\frac{\sin ^3 x}{3}-\frac{\sin ^5 x}{5}+c\)
- C \(\frac{\sin ^3 x}{3}+\frac{\sin ^5 x}{5}+c\)
- D \(\sin ^3 x+\sin ^5 x+c\)
Answer & Solution
Correct Answer
(B) \(\frac{\sin ^3 x}{3}-\frac{\sin ^5 x}{5}+c\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { Let } I=\int \cos ^3 x e^{\log (\sin x)^2} d x \\
& =\int \cos ^3 x(\sin x)^2 d x=\int \cos ^2 x(\sin x)^2 \cos x d x \\
& =\int \cos ^3 x(\sin x)^2 d x=\int \cos ^2 x(\sin x)^2 \cos x d x
\end{aligned}\)
Put \(\sin x=t \Rightarrow \cos x d x=d t\)
\(\therefore \mathrm{I}=\int\left(\mathrm{t}^2-\mathrm{t}^4\right) \mathrm{dt}=\frac{\mathrm{t}^3}{3}-\frac{\mathrm{t}^5}{5}+\mathrm{c}=\frac{\sin ^3 \mathrm{x}}{3}-\frac{\sin ^5 \mathrm{x}}{5}+\mathrm{c}\)
& \text { Let } I=\int \cos ^3 x e^{\log (\sin x)^2} d x \\
& =\int \cos ^3 x(\sin x)^2 d x=\int \cos ^2 x(\sin x)^2 \cos x d x \\
& =\int \cos ^3 x(\sin x)^2 d x=\int \cos ^2 x(\sin x)^2 \cos x d x
\end{aligned}\)
Put \(\sin x=t \Rightarrow \cos x d x=d t\)
\(\therefore \mathrm{I}=\int\left(\mathrm{t}^2-\mathrm{t}^4\right) \mathrm{dt}=\frac{\mathrm{t}^3}{3}-\frac{\mathrm{t}^5}{5}+\mathrm{c}=\frac{\sin ^3 \mathrm{x}}{3}-\frac{\sin ^5 \mathrm{x}}{5}+\mathrm{c}\)
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