int {frac{{{{sin }^8}x - {{cos }^8}x}}{{1 - 2 | vedclass

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

Question

(B) ધારો કે $I = \int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}\;dx}$.આપણે જાણીએ છીએ કે $1 = (\sin^2 x + \cos^2 x)^2 = \s

Vedclass · Accepted Answer

$-\frac{1}{2}\sin 2x + c$

Vedclass · Answer

(B) ધારો કે $I = \int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}\;dx}$.આપણે જાણીએ છીએ કે $1 = (\sin^2 x + \cos^2 x)^2 = \sin^4 x + \cos^4 x + 2\sin^2 x \cos^2 x$,તેથી છેદ $1 - 2\sin^2 x \cos^2 x = \sin^4 x + \cos^4 x$ થાય.આમ,$I = \int {\frac{{(\sin^4 x - \cos^4 x)(\sin^4 x + \cos^4 x)}}{{\sin^4 x + \cos^4 x}}\;dx}$.$I = \int {(\sin^4 x - \cos^4 x)\,dx}$.નિત્યસમ $a^2 - b^2 = (a-b)(a+b)$ નો ઉપયોગ કરતા,$\sin^4 x - \cos^4 x = (\sin^2 x - \cos^2 x)(\sin^2 x + \cos^2 x)$ મળે.કારણ કે $\sin^2 x + \cos^2 x = 1$,તેથી સંકલન $I = \int {(\sin^2 x - \cos^2 x)\,dx}$ બને.નિત્યસમ $\cos 2x = \cos^2 x - \sin^2 x$ નો ઉપયોગ કરતા,$\sin^2 x - \cos^2 x = -\cos 2x$ મળે.તેથી,$I = \int {-\cos 2x\,dx} = -\frac{\sin 2x}{2} + c$.

$\int {\frac{{{{\sin }^8}x - {{\cos }^8}x}}{{1 - 2{{\sin }^2}x{{\cos }^2}x}}\;dx} = $

Similar Questions

જો $k \in (1, \infty)$ હોય,તો $\int \frac{1}{1+k \cos x} d x=$

જો $\int \frac{3e^x - 5e^{-x}}{4e^x + 5e^{-x}} dx = px + q \cdot \log |4e^x + 5e^{-x}| + C$ હોય,તો

$\int(\sqrt{\tan x}+\sqrt{\cot x}) d x=$

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

$\begin{aligned} & \text{જો } 5(f(x))^2 = x f(x) + 30 \text{ અને } \\ & \int \frac{3 x^3 + (1 - 30 x^2) f(x)}{(10 f(x) - x)(x^3 - f(x))^2} dx \\ & = \frac{A}{B x^3 + D f(x)} + C, \text{ તો } A + B + D = \end{aligned}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module