જો int frac{cos x , dx}{sin ^{3} x left(1+sin | vedclass

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

Question

(A) ધારો કે $I = \int \frac{\cos x \, dx}{\sin ^{3} x \left(1+\sin ^{6} x\right)^{2 / 3}}$.$u = \sin x$ લેતા,$du = \cos x \, dx$ મળે.$I = \int \frac{d

Vedclass · Accepted Answer

$-2$

Vedclass · Answer

(A) ધારો કે $I = \int \frac{\cos x \, dx}{\sin ^{3} x \left(1+\sin ^{6} x\right)^{2 / 3}}$.$u = \sin x$ લેતા,$du = \cos x \, dx$ મળે.$I = \int \frac{du}{u^3 (1+u^6)^{2/3}}$.કૌંસમાંથી $u^6$ સામાન્ય લેતા: $I = \int \frac{du}{u^3 (u^6(\frac{1}{u^6}+1))^{2/3}} = \int \frac{du}{u^3 \cdot u^4 (\frac{1}{u^6}+1)^{2/3}} = \int \frac{du}{u^7 (u^{-6}+1)^{2/3}}$.$t = u^{-6}+1$ લેતા,$dt = -6u^{-7} \, du$,તેથી $u^{-7} \, du = -\frac{1}{6} \, dt$.$I = -\frac{1}{6} \int t^{-2/3} \, dt = -\frac{1}{6} \cdot \frac{t^{1/3}}{1/3} + c = -\frac{1}{2} t^{1/3} + c$.$t = u^{-6}+1 = \frac{1+u^6}{u^6}$ પાછું મૂકતા,$I = -\frac{1}{2} \left(\frac{1+u^6}{u^6}\right)^{1/3} + c = -\frac{1}{2} \frac{(1+\sin^6 x)^{1/3}}{\sin^2 x} + c$.$f(x)(1+\sin^6 x)^{1/\lambda} + c$ સાથે સરખાવતા,$\lambda = 3$ અને $f(x) = -\frac{1}{2 \sin^2 x}$ મળે.તેથી $\lambda f\left(\frac{\pi}{3}\right) = 3 \cdot \left(-\frac{1}{2 \sin^2(\pi/3)}\right) = 3 \cdot \left(-\frac{1}{2 \cdot (3/4)}\right) = 3 \cdot \left(-\frac{2}{3}\right) = -2$.

જો $\int \frac{\cos x \, dx}{\sin ^{3} x \left(1+\sin ^{6} x\right)^{2 / 3}} = f(x) \left(1+\sin ^{6} x\right)^{1 / \lambda} + c$ જ્યાં $c$ એ સંકલનનો અચળાંક છે,તો $\lambda f\left(\frac{\pi}{3}\right)$ ની કિંમત શોધો.

Similar Questions

$\int \frac{dx}{\sin^6 x + \cos^6 x} = $

$\int \frac{4 e^{x}+6 e^{-x}}{9 e^{x}-4 e^{-x}} d x=A x+B \log \left|9 e^{2 x}-4\right|+c$,હોય તો (જ્યાં $c$ એ સંકલનનો અચળાંક છે)

વિધેયનું સંકલન કરો: $\frac{x+2}{\sqrt{x^{2}+2 x+3}}$

જો $\int \frac{x^4+1}{x^6+1} dx = A \tan^{-1} x + B \tan^{-1} x^3 + c$ હોય,તો $(A, B) =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module