Search any question & find its solution
Question:
Answered & Verified by Expert
\(\int \sin ^3(x) \cdot \cos ^3(x) d x=\)
Options:
Solution:
1627 Upvotes
Verified Answer
The correct answer is:
\(\frac{1}{4} \sin ^4(x)-\frac{1}{6} \sin ^6(x)+c\)
\(\begin{aligned}
& \int \sin ^3 x \cdot \cos ^3 x d x \\
& \quad=\int \sin ^3 x \cdot \cos x\left(1-\sin ^2 x\right) d x
\end{aligned}\)
Let \(t=\sin x \Rightarrow d t=\cos x \cdot d x\)
\(\begin{aligned}
& \Rightarrow \quad \int t^3 \cdot\left(1-t^2\right) d t=\int\left(t^3-t^5\right) d t \\
& \Rightarrow \quad \frac{t^4}{4}-\frac{t^6}{6}+C=\frac{\sin ^4 x}{4}-\frac{\sin ^6 x}{6}+C
\end{aligned}\)
& \int \sin ^3 x \cdot \cos ^3 x d x \\
& \quad=\int \sin ^3 x \cdot \cos x\left(1-\sin ^2 x\right) d x
\end{aligned}\)
Let \(t=\sin x \Rightarrow d t=\cos x \cdot d x\)
\(\begin{aligned}
& \Rightarrow \quad \int t^3 \cdot\left(1-t^2\right) d t=\int\left(t^3-t^5\right) d t \\
& \Rightarrow \quad \frac{t^4}{4}-\frac{t^6}{6}+C=\frac{\sin ^4 x}{4}-\frac{\sin ^6 x}{6}+C
\end{aligned}\)
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.