KCET · Maths · Definite Integration
\(\int_{-\pi}^\pi\left(1-x^2\right) \sin x \cdot \cos ^2 x d x\) is
- A \(\pi-\frac{\pi^2}{3}\)
- B \(2 \pi-\pi^3\)
- C \(\pi-\frac{\pi^3}{2}\)
- D \(0\)
Answer & Solution
Correct Answer
(D) \(0\)
Step-by-step Solution
Detailed explanation
\(\because\) Let \(f(x)=\left(1-x^2\right) \sin x \cdot \cos ^2 x\)
\(f(-x)=\left(1-(-x)^2\right) \sin (-x) \cdot \cos ^2(-x)\)
\(=\left(1-x^2\right)(-\sin x)\left(\cos ^2 x\right)\)
\(=-\left(1-x^2\right)(\sin x)\left(\cos ^2 x\right)\)
So, \(\quad f(x)=-f(-x)\)
So, \(f(x)\) is an odd function.
Hence, \(\int_{-\pi}^\pi\left(1-x^2\right) \sin x \cdot \cos ^2 x d x=0\)
\(f(-x)=\left(1-(-x)^2\right) \sin (-x) \cdot \cos ^2(-x)\)
\(=\left(1-x^2\right)(-\sin x)\left(\cos ^2 x\right)\)
\(=-\left(1-x^2\right)(\sin x)\left(\cos ^2 x\right)\)
So, \(\quad f(x)=-f(-x)\)
So, \(f(x)\) is an odd function.
Hence, \(\int_{-\pi}^\pi\left(1-x^2\right) \sin x \cdot \cos ^2 x d x=0\)
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