The integral int_{pi /6}^{pi /4} {frac{{dx}}{ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Let $I = \int_{\pi /6}^{\pi /4} \frac{dx}{\sin 2x (	an^5 x + \cot^5 x)}$.Using $\sin 2x = \frac{2	an x}{1+	an^2 x}$,we have $\frac{1}{\sin 2x}

Vedclass · Accepted Answer

$\frac{1}{{10}}\left( {\frac{\pi }{4} - 	an ^{ - 1}\left( {\frac{1}{{9\sqrt 3 }}} ight)} ight)$

Vedclass · Answer

(B) Let $I = \int_{\pi /6}^{\pi /4} \frac{dx}{\sin 2x (	an^5 x + \cot^5 x)}$.Using $\sin 2x = \frac{2	an x}{1+	an^2 x}$,we have $\frac{1}{\sin 2x} = \frac{1+	an^2 x}{2	an x}$.$I = \int_{\pi /6}^{\pi /4} \frac{(1+	an^2 x) dx}{2	an x (	an^5 x + \cot^5 x)}$.Let $t = 	an x$,then $dt = \sec^2 x dx = (1+	an^2 x) dx$.When $x = \pi/6, t = 1/\sqrt{3}$. When $x = \pi/4, t = 1$.$I = \int_{1/\sqrt{3}}^{1} \frac{dt}{2t(t^5 + 1/t^5)} = \int_{1/\sqrt{3}}^{1} \frac{t^4 dt}{2(t^{10} + 1)}$.Let $u = t^5$,then $du = 5t^4 dt$,so $t^4 dt = du/5$.When $t = 1/\sqrt{3}, u = (1/\sqrt{3})^5 = 1/(9\sqrt{3})$. When $t = 1, u = 1$.$I = \frac{1}{2} \int_{1/(9\sqrt{3})}^{1} \frac{du/5}{u^2 + 1} = \frac{1}{10} [	an^{-1} u]_{1/(9\sqrt{3})}^{1}$.$I = \frac{1}{10} (	an^{-1}(1) - 	an^{-1}(1/(9\sqrt{3}))) = \frac{1}{10} (\frac{\pi}{4} - 	an^{-1}(\frac{1}{9\sqrt{3}}))$.

The integral $\int_{\pi /6}^{\pi /4} {\frac{{dx}}{{\sin 2x\left( {{{\tan }^5}x + {{\cot }^5}x} \right)}}} $ equals

Similar Questions

If $f(x) = \begin{cases} 2x^2 + 1, & x \leq 1 \\ 4x^3 - 1, & x > 1 \end{cases}$,then $\int_{0}^{2} f(x) dx$ is

If $f(x) = e^{-x} + 2e^{-2x} + 3e^{-3x} + \dots + \infty$,then $\int_{\ln 2}^{\ln 3} f(x) \, dx =$

The value of $x$ that satisfies the equation $\int_{\sqrt{2}}^x \frac{dt}{|t| \sqrt{t^2-1}} = \frac{\pi}{12}$ is

$\int_{-3}^3 |2-x| dx =$

$\int_0^\pi \frac{1}{4+3 \cos x} d x=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module