The value of the integral ∫0π/2(3√(cos⁡θ))/(cos⁡θ+sin⁡θ^5)dθ equals

Correct answer: 0.5. I=∫0(π)/2(3√(cos⁡θ))/(sin⁡θ+cos⁡θ^5).dθ …(i) I=∫0(π)/2(3√(sin⁡θ))/(cos⁡θ+sin⁡θ^5).dθ …(ii) using ∫abfx.dx=∫abfa+b-x.dx (i) + (ii) ⇒2I=∫0(π)/2(3√(sin⁡θ)+√(cos⁡θ))/(sin⁡θ+cos⁡θ^5).dθ ⇒2I=3∫0(π)/2(dθ)/(sin⁡θ+cos⁡θ^4) ⇒2I/3=∫0(π)/2(sec^2⁡θ.dθ)/(1+tan⁡θ^4) Put tan⁡θ=t^2 sec^2⁡θ.dθ=2t.dt when θ→0, t→0 θ→(π)/2, t→∞ ⇒2I/3=∫0∞2tdt/(1+t^4) ⇒I/3=∫0∞(t+1-11+t^4.dtI)/3=∫0∞1/(1+t^3)-1/(1+t^4).dt I/3=-1/(21+t^2)+1/(31+t^3)0∞ I/3=0+0--1/2+1/3 I/3=1/6⇒I=1/2

Q31 / 54 · JEE Advanced 2019 (Paper 2) · Mathematics

The value of the integral $\int_{0}^{π / 2} \frac{3 \sqrt{\cos θ}}{{(\sqrt{\cos θ} + \sqrt{\sin θ})}^{5}} d θ$ equals

Correct answer: 0.5

Solution

I = \int_{0}^{\frac{π}{2}} \frac{3 \sqrt{\cos θ}}{{(\sqrt{\sin θ} + \sqrt{\cos θ})}^{5}} . d θ

…(i)

I = \int_{0}^{\frac{π}{2}} \frac{3 \sqrt{\sin θ}}{{(\sqrt{\cos θ} + \sqrt{\sin θ})}^{5}} . d θ

…(ii)

using \int_{a}^{b} f (x) . d x = \int_{a}^{b} f (a + b - x) . d x

(i)

+

(ii)

\Rightarrow 2 I = \int_{0}^{\frac{π}{2}} \frac{3 (\sqrt{\sin θ} + \sqrt{\cos θ})}{{(\sqrt{\sin θ} + \sqrt{\cos θ})}^{5}} . d θ

\Rightarrow 2 I = 3 \int_{0}^{\frac{π}{2}} \frac{d θ}{{(\sqrt{\sin θ} + \sqrt{\cos θ})}^{4}}

\Rightarrow \frac{2 I}{3} = \int_{0}^{\frac{π}{2}} \frac{\sec^{2} θ . d θ}{{(1 + \sqrt{\tan θ})}^{4}}

Put

\tan θ = t^{2}

\sec^{2} θ . d θ = 2 t . d t

when

\begin{matrix} θ \to 0, & t \to 0 \end{matrix}

\begin{matrix} θ \to \frac{π}{2}, & t \to \infty \end{matrix}

\Rightarrow \frac{2 I}{3} = \int_{0}^{\infty} \frac{2 t d t}{{(1 + t)}^{4}}

\Rightarrow \frac{I}{3} = \int_{0}^{\infty} \frac{(t + 1) - 1}{{(1 + t)}^{4}} . d t

\frac{I}{3} = \int_{0}^{\infty} \frac{1}{{(1 + t)}^{3}} - \frac{1}{{(1 + t)}^{4}} . d t

\frac{I}{3} = {[- \frac{1}{2 {(1 + t)}^{2}} + \frac{1}{3 {(1 + t)}^{3}}]}_{0}^{\infty}

\frac{I}{3} = [0 + 0 - (- \frac{1}{2} + \frac{1}{3})]

\frac{I}{3} = \frac{1}{6} \Rightarrow I = \frac{1}{2}

Hence, the correct answer is 0.5.

The value of the integral ∫0π/23cos⁡θcos⁡θ+sin⁡θ5dθ equals

Solution

Relevant Questions

The value of the integral $\int_{0}^{π / 2} \frac{3 \sqrt{\cos θ}}{{(\sqrt{\cos θ} + \sqrt{\sin θ})}^{5}} d θ$ equals