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Find the value of \(\rm \int_{0}^{\pi/2 } cos^3 x dx\)
1. \(\rm \frac{2}{3}\)
2. 1
3. 2
4. 3

1 Answer

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Best answer
Correct Answer - Option 1 : \(\rm \frac{2}{3}\)

Concept:

sin2 x + cos2 x = 1

\(\rm \frac{d(sinx)}{dx} = cos x\)

\(\rm \int x^ndx = \frac{x^{n + 1}}{n + 1}\)

\(\rm \int dx = x\)

Calculation:

I = \(\rm \int_{0}^{\pi/2 } cos^3 x dx\)

\(\rm \int_{0}^{\pi/2 } cos^2 x .cos x dx\)

\(\rm \int_{0}^{\pi/2 } (1 - sin^2 x) .cos x dx\)

Let, sin x = u

cos x dx = du

\(\rm \int_{0}^{1} (1 - u^2)du\)

\(\rm \left [u - \frac{u^{3}}{3} \right ]_{0}^{1}\)

= 1 - 0 - \(\rm \frac{1}{3} + 0\)

\(\rm \frac{2}{3}\)

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