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  • Evaluate the integrals in Exercises 1 to 8 using substitution. 2

Evaluate the integrals in Exercises 1 to 8 using substitution.

    Q2.    \int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi

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\int^\frac{\pi}{2}_0\sqrt{\sin\phi}\cos^5\phi d\phi
let \sin \phi = t \Rightarrow \cos \phi d\phi = dt
when \phi =0,t\rightarrow 0 and \phi =\pi/2,t\rightarrow 1

using the above substitution we can evaluate the integral as

\\\therefore \int_{0}^{1}\sqrt{t}(1-t^2)dt\\ =\int_{0}^{1} t^\frac{1}{2}(1+t^4-2t^2)dt\\ =\int_{0}^{1}t^\frac{1}{2}dt+\int_{0}^{1}t^{9/2}dt-2\int_{0}^{1}t^{5/2}dt\\ =[2t^{3/2}/3+2t^{11/2}/11+4t^{7/2}/7]^1_0\\ =\frac{64}{231}

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manish

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