← 2015 Paper 1

UPSC 2015 Maths Optional Paper 1 Q1d — Step-by-Step Solution

10 marks · Section A

Indefinite integrals · Calculus · asked 7× in 13 yrs · Read the full method →

Question

Evaluate the following integral:

I=∫π/6π/3sin⁡x3sin⁡x3+cos⁡x3 dx.I=\int_{\pi/6}^{\pi/3}\dfrac{\sqrt[3]{\sin x}}{\sqrt[3]{\sin x}+\sqrt[3]{\cos x}}\,dx.

Technique

King’s reflection property f(x)+f(a+b−x)f(x)+f(a+b-x) exploiting sin⁡↔cos⁡\sin\leftrightarrow\cos symmetry about x=π/4x=\pi/4.

Solution

Strategy. Use King’s property ∫abf(x) dx=∫abf(a+b−x) dx\int_a^b f(x)\,dx=\int_a^b f(a+b-x)\,dx with a+b=π/2a+b=\pi/2. Under x↦π/2−xx\mapsto\pi/2-x, sin⁡x↔cos⁡x\sin x\leftrightarrow\cos x.

Step 1 — Apply the substitution x→π/2−xx\to\pi/2-x

I=∫π/6π/3sin⁡(π/2−x)3sin⁡(π/2−x)3+cos⁡(π/2−x)3 dx=∫π/6π/3cos⁡x3cos⁡x3+sin⁡x3 dx.I=\int_{\pi/6}^{\pi/3}\dfrac{\sqrt[3]{\sin(\pi/2-x)}}{\sqrt[3]{\sin(\pi/2-x)}+\sqrt[3]{\cos(\pi/2-x)}}\,dx=\int_{\pi/6}^{\pi/3}\dfrac{\sqrt[3]{\cos x}}{\sqrt[3]{\cos x}+\sqrt[3]{\sin x}}\,dx.

Step 2 — Add the two expressions for II

2I=∫π/6π/3sin⁡x3+cos⁡x3sin⁡x3+cos⁡x3 dx=∫π/6π/31 dx=π3−π6=π6.2I=\int_{\pi/6}^{\pi/3}\dfrac{\sqrt[3]{\sin x}+\sqrt[3]{\cos x}}{\sqrt[3]{\sin x}+\sqrt[3]{\cos x}}\,dx=\int_{\pi/6}^{\pi/3}1\,dx=\dfrac{\pi}{3}-\dfrac{\pi}{6}=\dfrac{\pi}{6}.

Step 3 — Solve

Answer

  I=π12.  \boxed{\;I=\dfrac{\pi}{12}.\;}
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