Login

Integrate wrt x- (sin^6x + cos^6x )/sin^2x cos^2x

$\int \frac{\sin^{6} x + \cos^{6} x}{\sin^{2} x \cos^{2} x} ⅆ x = \int \frac{{(\sin^{2} x)}^{3} + {(\cos^{2} x)}^{3}}{\sin^{2} x \cos^{2} x} ⅆ x = \int \frac{{(\sin^{2} x + \cos^{2} x)}^{3} - 3 \sin^{2} x \cos^{2} x (\sin^{2} x + \cos^{2} x)}{\sin^{2} x \cos^{2} x} d x ({(a + b)}^{3} = a^{3} + b^{3} + 3 a b (a + b)) = \int \frac{{(1)}^{3} - 3 \sin^{2} x \cos^{2} x (1)}{\sin^{2} x \cos^{2} x} d x = \int \frac{1 - 3 \sin^{2} x \cos^{2} x}{\sin^{2} x \cos^{2} x} d x = \int \frac{1}{\sin^{2} x \cos^{2} x} d x - \int \frac{3 \sin^{2} x \cos^{2} x}{\sin^{2} x \cos^{2} x} d x = \int \frac{\sin^{2} x + \cos^{2} x}{\sin^{2} x \cos^{2} x} d x - \int 3 d x = \int (\sec^{2} x + {\cos e c}^{2} x) d x - 3 x = \int \sec^{2} x d x + \int {\cos e c}^{2} x d x - 3 x = \tan x - \cot x - 3 x + C H e n c e, \int \frac{\sin^{6} x + \cos^{6} x}{\sin^{2} x \cos^{2} x} ⅆ x = \tan x - \cot x - 3 x + C$

228

View Full Answer

Copyright © 2024 Aakash EduTech Pvt. Ltd. All rights reserved.

Aakash EduTech Pvt. Ltd.

What are you looking for?