જો int(sin x )^{frac{-11}{2}}(cos x )^{frac{- | vedclass

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

Question

(C) આપેલ સંકલન $I = \int (\sin x)^{-\frac{11}{2}} (\cos x)^{-\frac{5}{2}} dx = \int (	an x)^{-\frac{11}{2}} \sec^8 x dx$ તરીકે લેતા. $\sec^8 x = (1 +

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) આપેલ સંકલન $I = \int (\sin x)^{-\frac{11}{2}} (\cos x)^{-\frac{5}{2}} dx = \int (	an x)^{-\frac{11}{2}} \sec^8 x dx$ તરીકે લેતા. $\sec^8 x = (1 + 	an^2 x)^3 \sec^2 x$ હોવાથી, $	an x = t$ લેતા $dt = \sec^2 x dx$ મળે. $I = \int t^{-\frac{11}{2}} (1 + t^2)^3 dt = \int t^{-\frac{11}{2}} (1 + t^6 + 3t^4 + 3t^2) dt = \int (t^{-\frac{11}{2}} + t^{\frac{1}{2}} + 3t^{-\frac{3}{2}} + 3t^{-\frac{7}{2}}) dt$. સંકલન કરતા: $-\frac{2}{9} t^{-\frac{9}{2}} - \frac{6}{5} t^{-\frac{5}{2}} - 6 t^{-\frac{1}{2}} + \frac{2}{3} t^{\frac{3}{2}} + C$. અહીં $p_1=2, q_1=9, p_2=6, q_2=5, p_3=6, q_3=1, p_4=2, q_4=3$ મળે છે. તેથી $\frac{15 \cdot 2 \cdot 6 \cdot 6 \cdot 2}{9 \cdot 5 \cdot 1 \cdot 3} = 16$.

Similar Questions

$\int \sqrt{x^2 - 8x + 7} \, dx = $

જો $\int \sec ^2 x \operatorname{cosec}^4 x \, dx = -\frac{1}{3} \cot ^3 x + k \tan x - 2 \cot x + C$ હોય,તો $k$ ની કિંમત શોધો.

$\int \frac{x^{2}}{(x \sin x+\cos x)^{2}} d x$ ની કિંમત શોધો.

$\int \frac{dx}{(1+\sqrt{x}) \sqrt{x-x^2}} = $

જો $\int \sqrt{\sec 2x - 1} \, dx = \alpha \log_e \left| \cos 2x + \beta + \sqrt{\cos 2x (1 + \cos \frac{1}{\beta} x)} \right| + C$ હોય,તો $\beta - \alpha$ ની કિંમત શોધો.

Vedclass Test Series

Exam Paper Generator

Online Exam Module