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 $I_n = \int \frac{t^{2n}}{1+t^2} dt$ હોય,તો $I_{n+1} =$

$\int \sqrt{x^2-6x-16} \, dx$ ની કિંમત શોધો.

જો $\int(\sqrt{\operatorname{cosec} x+1}) d x=k \tan ^{-1}(f(x))+c$ હોય, તો $\frac{1}{k} f\left(\frac{\pi}{6}\right)=$

જો $\int \frac{(\sqrt{1+x^2}+x)^{10}}{(\sqrt{1+x^2}-x)^9} dx = \frac{1}{m}((\sqrt{1+x^2}+x)^n (n\sqrt{1+x^2}-x)) + C$,જ્યાં $C$ એ સંકલનનો અચળાંક છે અને $m, n \in N$,તો $m+n$ ની કિંમત શોધો.

$\int \frac{d x}{(x-1)^{\frac{3}{4}}(x+2)^{\frac{5}{4}}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module