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Miscellaneous Exercise 4 · Q62

Q.Evaluate: ∫π/4π/2cos⁡θ[cos⁡θ2+sin⁡θ2]3 dθ\int_{\pi/4}^{\pi/2} \dfrac{\cos\theta}{\left[\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\right]^3}\,d\theta

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[cos⁡θ2+sin⁡θ2]2=1+2sin⁡θ2cos⁡θ2=1+sin⁡θ\big[\cos\tfrac\theta2+\sin\tfrac\theta2\big]^2=1+2\sin\tfrac\theta2\cos\tfrac\theta2=1+\sin\theta, and since θ/2∈[π/8,π/4]\theta/2\in[\pi/8,\pi/4] both terms are positive, so [cos⁡θ2+sin⁡θ2]3=(1+sin⁡θ)3/2\big[\cos\tfrac\theta2+\sin\tfrac\theta2\big]^3=(1+\sin\theta)^{3/2}.

Let u=1+sin⁡θ, du=cos⁡θ dθu=1+\sin\theta,\ du=\cos\theta\,d\theta; limits θ=π/4→u=1+22\theta=\pi/4\to u=1+\tfrac{\sqrt2}2, θ=π/2→u=2\theta=\pi/2\to u=2.

∫u−3/2 du=−2u−1/2=−2u.\int u^{-3/2}\,du=-2u^{-1/2}=-\frac2{\sqrt u}. …

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