int frac{cos^3 x + cos^5 x}{sin^2 x + sin^4 x | vedclass

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

Question

(C) ધારો કે $I = \int \frac{\cos^3 x + \cos^5 x}{\sin^2 x + \sin^4 x} dx$. $= \int \frac{\cos^3 x (1 + \cos^2 x)}{\sin^2 x (1 + \sin^2 x)} dx$ $= \int

Vedclass · Accepted Answer

$\sin x - 2(\sin x)^{-1} - 6 	an^{-1}(\sin x) + c$

Vedclass · Answer

(C) ધારો કે $I = \int \frac{\cos^3 x + \cos^5 x}{\sin^2 x + \sin^4 x} dx$. $= \int \frac{\cos^3 x (1 + \cos^2 x)}{\sin^2 x (1 + \sin^2 x)} dx$ $= \int \frac{\cos^2 x (1 + \cos^2 x)}{\sin^2 x (1 + \sin^2 x)} \cos x dx$ $= \int \frac{(1 - \sin^2 x) (1 + 1 - \sin^2 x)}{\sin^2 x (1 + \sin^2 x)} \cos x dx$ $= \int \frac{(1 - \sin^2 x) (2 - \sin^2 x)}{\sin^2 x (1 + \sin^2 x)} \cos x dx$. ધારો કે $\sin x = t$,તેથી $\cos x dx = dt$. $I = \int \frac{(1 - t^2) (2 - t^2)}{t^2 (1 + t^2)} dt$ $= \int \frac{t^4 - 3t^2 + 2}{t^2 (1 + t^2)} dt$ $= \int \frac{t^2(t^2 + 1) - 4t^2 + 2}{t^2 (1 + t^2)} dt$ $= \int \left( 1 + \frac{2 - 4t^2}{t^2 (1 + t^2)} ight) dt$ $= \int 1 dt + \int \left( \frac{2}{t^2} - \frac{6}{1 + t^2} ight) dt$ $= t - \frac{2}{t} - 6 	an^{-1} t + c$ $= \sin x - 2(\sin x)^{-1} - 6 	an^{-1} (\sin x) + c$.

$\int \frac{\cos^3 x + \cos^5 x}{\sin^2 x + \sin^4 x} dx =$

Similar Questions

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.

$\int \frac{\sin x+8 \cos x}{4 \sin x+6 \cos x} d x$ ની કિંમત શોધો.

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

$\int \frac{(1-4 \sin^2 x) \cos x}{\cos (3x+2)} dx =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module