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Definite Integration question

2019 · 9 Apr · Shift 1 · Q27
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  5. /2019 · 9 Apr · Shift 1 · Q27

Definite Integration question

2019 · 9 Apr · Shift 1 · Q27

JEE MainMathematicsDefinite IntegrationMCQ+4 / −1
The value of ∫0π/2sin⁡3xsin⁡x+cos⁡xdx\int\limits_0^{\pi /2} {{{{{\sin }^3}x} \over {\sin x + \cos x}}dx}0∫π/2​sinx+cosxsin3x​dx is
  1. A
    π−28{{\pi - 2} \over 8}8π−2​
  2. B
    π−24{{\pi - 2} \over 4}4π−2​
  3. C
    π−12{{\pi - 1} \over 2}2π−1​
  4. D
    π−14{{\pi - 1} \over 4}4π−1​
View written solutionFree

Correct answer: D

  1. Let I=∫0π/2sin⁡3xsin⁡x+cos⁡x dx.I=\int_0^{\pi/2}\frac{\sin^3 x}{\sin x+\cos x}\,dx.I=∫0π/2​sinx+cosxsin3x​dx.

We use the standard property for integrals over [0,π/2][0,\pi/2][0,π/2]: If I=∫0π/2f(sin⁡x,cos⁡x) dx,I=\int_0^{\pi/2} f(\sin x,\cos x)\,dx,I=∫0π/2​f(sinx,cosx)dx, then under the substitution x↦π2−xx\mapsto \frac{\pi}{2}-xx↦2π​−x, I=∫0π/2f(cos⁡x,sin⁡x) dx.I=\int_0^{\pi/2} f(\cos x,\sin x)\,dx.I=∫0π/2​f(cosx,sinx)dx.

So, I=∫0π/2cos⁡3xsin⁡x+cos⁡x dx.I=\int_0^{\pi/2}\frac{\cos^3 x}{\sin x+\cos x}\,dx.I=∫0π/2​sinx+cosxcos3x​dx.

  1. Add the two expressions for III: 2I=∫0π/2sin⁡3x+cos⁡3xsin⁡x+cos⁡x dx.2I=\int_0^{\pi/2}\frac{\sin^3 x+\cos^3 x}{\sin x+\cos x}\,dx.2I=∫0π/2​sinx+cosxsin3x+cos3x​dx.

Now use a3+b3=(a+b)(a2−ab+b2).a^3+b^3=(a+b)(a^2-ab+b^2).a3+b3=(a+b)(a2−ab+b2). Hence, sin⁡3x+cos⁡3xsin⁡x+cos⁡x=sin⁡2x−sin⁡xcos⁡x+cos⁡2x=1−sin⁡xcos⁡x.\frac{\sin^3 x+\cos^3 x}{\sin x+\cos x}=\sin^2 x-\sin x\cos x+\cos^2 x=1-\sin x\cos x.sinx+cosxsin3x+cos3x​=sin2x−sinxcosx+cos2x=1−sinxcosx.

Therefore, 2I=∫0π/2(1−sin⁡xcos⁡x) dx.2I=\int_0^{\pi/2}(1-\sin x\cos x)\,dx.2I=∫0π/2​(1−sinxcosx)dx.

  1. Evaluate the integral: ∫0π/21 dx=π2.\int_0^{\pi/2}1\,dx=\frac{\pi}{2}.∫0π/2​1dx=2π​.

Also, ∫0π/2sin⁡xcos⁡x dx=12∫0π/2sin⁡2x dx.\int_0^{\pi/2}\sin x\cos x\,dx=\frac12\int_0^{\pi/2}\sin 2x\,dx.∫0π/2​sinxcosxdx=21​∫0π/2​sin2xdx.

Now,

=\frac12\left[-\frac{\cos 2x}{2}\right]_0^{\pi/2} =\frac12\left(\frac{1}{2}-\left(-\frac{1}{2}\right)?\right).$$ Let us do it carefully: $$\int_0^{\pi/2}\sin 2x\,dx=\left[-\frac{\cos 2x}{2}\right]_0^{\pi/2} =\left(-\frac{\cos\pi}{2}\right)-\left(-\frac{\cos0}{2}\right) =\frac12-\left(-\frac12\right)=1.$$ So, $$\int_0^{\pi/2}\sin x\cos x\,dx=\frac12.$$ Thus, $$2I=\frac{\pi}{2}-\frac12=\frac{\pi-1}{2}.$$ Hence, $$I=\frac{\pi-1}{4}.$$ 4. Compare with the options: - A: $\frac{\pi-2}{8}$ - B: $\frac{\pi-2}{4}$ - C: $\frac{\pi-1}{2}$ - D: $\frac{\pi-1}{4}$ So the correct option is **D**.
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