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NCERT Class XII Mathematics Chapter - - Solutions

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Exercise - 7.10
Evaluate the definite integrals in Exercises 198 to 205 using substitution.
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Question : 199 of 271
Marks: +1, -0
∫0π/2sin⁡ϕcos⁡5ϕ dϕ\int\limits_{0}^{\pi/2} \sqrt{\sin\phi} \cos^5\phi \, d\phi
Solution:  
Let I=∫0π/2sin⁡ϕcos⁡5ϕ dϕI = \int\limits_{0}^{\pi/2} \sqrt{\sin\phi} \cos^5\phi \, d\phi
=∫0π/2sin⁡ϕcos⁡4ϕcos⁡ϕ dϕ= \int\limits_{0}^{\pi/2} \sqrt{\sin\phi} \cos^4\phi \cos\phi \, d\phi
=∫0π/2sin⁡ϕ(1−sin⁡2ϕ)2cos⁡ϕ dϕ= \int\limits_{0}^{\pi/2} \sqrt{\sin\phi} (1-\sin^2\phi)^2 \cos\phi \, d\phi
Put sin⁡ϕ=t⇒cos⁡ϕ dϕ=dt\sin\phi = t \Rightarrow \cos\phi \, d\phi = dt
When ϕ=0,t=0\phi = 0, t = 0 ; when ϕ=π/2,t=1\phi = \pi/2, t = 1
∴I=∫01t(1−t2)2 dt\therefore I = \int\limits_{0}^{1} \sqrt{t} (1-t^2)^2 \, dt
=∫01t(1−2t2+t4) dt= \int\limits_{0}^{1} \sqrt{t} (1 - 2t^2 + t^4) \, dt
=∫01(t1/2+t9/2−2t5/2) dt= \int\limits_{0}^{1} (t^{1/2} + t^{9/2} - 2t^{5/2}) \, dt
=[23t3/2+211t11/2−47t7/2]01= \left[ \frac{2}{3} t^{3/2} + \frac{2}{11} t^{11/2} - \frac{4}{7} t^{7/2} \right]_{0}^{1}
=23+211−47= \frac{2}{3} + \frac{2}{11} - \frac{4}{7}
=154+42−1323×11×7= \frac{154 + 42 - 132}{3 \times 11 \times 7}
=64231= \frac{64}{231}
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