Question:
The value of the integral $\int\limits^{\pi/2}_{0}\sin^{5}\, x\,dx$ is
The value of the integral $\int\limits^{\pi/2}_{0}\sin^{5}\, x\,dx$ is
Updated On: Aug 9, 2024
- $\frac{4}{15}$
- $\frac{8}{5}$
- $\frac{8}{15}$
- $\frac{4}{5}$
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The Correct Option is C
Solution and Explanation
$I=\int\limits_{0}^{\pi / 2} \sin ^{4} x \cdot \sin x \,d x$
$=\int\limits_{0}^{\pi / 2}\left(1-\cos ^{2} x\right)^{2} \sin x\, d x$
Put $\cos x=t$
$\Rightarrow -\sin x\, d x=d t$
$=-\int\limits_{1}^{0}\left(1-t^{2}\right)^{2} d t=\int\limits_{0}^{1}\left(t^{4}-2 t^{2}+1\right) d t$
$\left[\because-\int_{a}^{b} f(x) d x=\int\limits_{a}^{b} f(x) d x\right]$
$=\frac{1}{5}\left(t^{5}\right)_{0}^{1}-\frac{2}{3}\left(t^{3}\right)_{0}^{1}+(t)_{0}^{1}$
$=\frac{1}{5}-\frac{2}{3}+1=\frac{3-10+15}{15}=\frac{8}{15}$
$=\int\limits_{0}^{\pi / 2}\left(1-\cos ^{2} x\right)^{2} \sin x\, d x$
Put $\cos x=t$
$\Rightarrow -\sin x\, d x=d t$
$=-\int\limits_{1}^{0}\left(1-t^{2}\right)^{2} d t=\int\limits_{0}^{1}\left(t^{4}-2 t^{2}+1\right) d t$
$\left[\because-\int_{a}^{b} f(x) d x=\int\limits_{a}^{b} f(x) d x\right]$
$=\frac{1}{5}\left(t^{5}\right)_{0}^{1}-\frac{2}{3}\left(t^{3}\right)_{0}^{1}+(t)_{0}^{1}$
$=\frac{1}{5}-\frac{2}{3}+1=\frac{3-10+15}{15}=\frac{8}{15}$
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Concepts Used:
Integrals of Some Particular Functions
There are many important integration formulas which are applied to integrate many other standard integrals. In this article, we will take a look at the integrals of these particular functions and see how they are used in several other standard integrals.
Integrals of Some Particular Functions:
- ∫1/(x2 – a2) dx = (1/2a) log|(x – a)/(x + a)| + C
- ∫1/(a2 – x2) dx = (1/2a) log|(a + x)/(a – x)| + C
- ∫1/(x2 + a2) dx = (1/a) tan-1(x/a) + C
- ∫1/√(x2 – a2) dx = log|x + √(x2 – a2)| + C
- ∫1/√(a2 – x2) dx = sin-1(x/a) + C
- ∫1/√(x2 + a2) dx = log|x + √(x2 + a2)| + C
These are tabulated below along with the meaning of each part.
