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Exercise 7.9 · Q2

Q.Evaluate the integral using substitution ∫0π/2sin⁡ϕ cos⁡5ϕ dϕ\int_{0}^{\pi/2}\sqrt{\sin\phi}\,\cos^5\phi\,d\phi

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✓ Free question

Substitute t=sin⁡ϕt=\sin\phi; the odd cos⁡\cos power supplies dtdt and the integral becomes a simple polynomial, giving 64231\dfrac{64}{231}.

Choosing the substitution

Because cos⁡ϕ\cos\phi appears to an odd power, we can peel off one factor to serve as dtdt and convert the even remainder to sin⁡ϕ\sin\phi. Set t=sin⁡ϕt=\sin\phi, so dt=cos⁡ϕ dϕdt=\cos\phi\,d\phi.

1. Rewrite the integrand

cos⁡5ϕ dϕ=cos⁡4ϕ⋅cos⁡ϕ dϕ=(1−sin⁡2ϕ)2cos⁡ϕ dϕ=(1−t2)2 dt.\cos^5\phi\,d\phi=\cos^4\phi\cdot\cos\phi\,d\phi=(1-\sin^2\phi)^2\cos\phi\,d\phi=(1-t^2)^2\,dt.

The limits change as ϕ:0→π2\phi:0\to\tfrac\pi2 gives t:0→1t:0\to1, so

∫0π/2sin⁡ϕ cos⁡5ϕ dϕ=∫01t1/2(1−t2)2 dt.\int_0^{\pi/2}\sqrt{\sin\phi}\,\cos^5\phi\,d\phi=\int_0^1 t^{1/2}(1-t^2)^2\,dt.

2. Expand and integrate

t1/2(1−t2)2=t1/2(1−2t2+t4)=t1/2−2t5/2+t9/2,t^{1/2}(1-t^2)^2=t^{1/2}(1-2t^2+t^4)=t^{1/2}-2t^{5/2}+t^{9/2},

∫01(t1/2−2t5/2+t9/2) dt=[23t3/2−47t7/2+211t11/2]01=23−47+211.\int_0^1\big(t^{1/2}-2t^{5/2}+t^{9/2}\big)\,dt=\Big[\tfrac23 t^{3/2}-\tfrac47 t^{7/2}+\tfrac2{11}t^{11/2}\Big]_0^1=\frac23-\frac47+\frac2{11}.

3. Add the fractions

Common denominator 231231:

154231−132231+42231=64231.\frac{154}{231}-\frac{132}{231}+\frac{42}{231}=\frac{64}{231}.

✓Final answer

∫0π/2sin⁡ϕ cos⁡5ϕ dϕ=64231\displaystyle\int_0^{\pi/2}\sqrt{\sin\phi}\,\cos^5\phi\,d\phi=\frac{64}{231}

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