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JEE Advanced 2019Paper 2
Exam
JEE Advanced
Year
2019
Paper
Paper 2
Subject
Mathematics
Chapter
Definite Integration

JEE Advanced 2019 · Paper 2 · Mathematics · Definite Integration

The value of the integral ∫ _ 0 ^ π / 2 3 cos θ ( cos θ + sin θ ) ^...

The value of the integral ∫0π/23cosθ(cosθ+sinθ)5dθ\mathop{∫}_{0}^{\pi / 2}\dfrac{3\sqrt{cos \theta}}{{\left(\sqrt{cos \theta} + \sqrt{sin \theta}\right)}^{5}}d\theta equals

Correct answer

0

Step-by-step solution

I=∫0π23cosθ(sinθ+cosθ)5.dθI = \mathop{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{3\sqrt{cos \theta}}{{\left(\sqrt{sin \theta} + \sqrt{cos \theta}\right)}^{5}} . d\theta …(i)
I=∫0π23sinθ(cosθ+sinθ)5.dθI = \mathop{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{3\sqrt{sin \theta}}{{\left(\sqrt{cos \theta} + \sqrt{sin \theta}\right)}^{5}} . d\theta …(ii) using∫abf(x).dx=∫abf(a+b−x).dxusing\mathop{∫}_{a}^{b}f\left(x\right) . dx = \mathop{∫}_{a}^{b}f\left(a + b - x\right) . dx
(i) ++ (ii)
⇒2I=∫0π23(sinθ+cosθ)(sinθ+cosθ)5.dθ⇒ 2I = \mathop{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{3\left(\sqrt{sin \theta} + \sqrt{cos \theta}\right)}{{\left(\sqrt{sin \theta} + \sqrt{cos \theta}\right)}^{5}} . d\theta
⇒2I=3∫0π2dθ(sinθ+cosθ)4⇒ 2I = 3\mathop{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{d\theta}{{\left(\sqrt{sin \theta} + \sqrt{cos \theta}\right)}^{4}}
⇒2I3=∫0π2sec⁡2θ.dθ(1+tanθ)4⇒ \dfrac{2I}{3} = \mathop{∫}_{0}^{\dfrac{\pi}{2}}\dfrac{\sec^{2} \theta .d\theta}{{\left(1 + \sqrt{tan \theta}\right)}^{4}}
Put tanθ=t2tan \theta = {t}^{2}
sec⁡2θ.dθ=2t.dt\sec^{2} \theta . d\theta = 2t . dt
when
{θ→0,t→0\begin{cases}\theta \to 0, & t \to 0\end{cases}
{θ→π2,t→∞\begin{cases}\theta \to \dfrac{\pi}{2}, & t \to \infty\end{cases}
⇒2I3=∫0∞2tdt(1+t)4⇒ \dfrac{2I}{3} = \mathop{∫}_{0}^{\infty}\dfrac{2tdt}{{\left(1 + t\right)}^{4}}
⇒I3=∫0∞(t+1)−1(1+t)4.dt⇒ \dfrac{I}{3} = \mathop{∫}_{0}^{\infty}\dfrac{\left(t + 1\right) - 1}{{\left(1 + t\right)}^{4}} . dt
I3=∫0∞1(1+t)3−1(1+t)4.dt\dfrac{I}{3} = \mathop{∫}_{0}^{\infty}\dfrac{1}{{\left(1 + t\right)}^{3}} - \dfrac{1}{{\left(1 + t\right)}^{4}} . dt
I3=[−12(1+t)2+13(1+t)3]0∞\dfrac{I}{3} = {\left[- \dfrac{1}{2{\left(1 + t\right)}^{2}} + \dfrac{1}{3{\left(1 + t\right)}^{3}}\right]}_{0}^{\infty}
I3=[0+0−(−12+13)]\dfrac{I}{3} = \left[0 + 0 - \left(- \dfrac{1}{2} + \dfrac{1}{3}\right)\right]
I3=16⇒I=12\dfrac{I}{3} = \dfrac{1}{6} ⇒ I = \dfrac{1}{2}
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