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