-sin 6 x-cos 6 x-sin n x cos 2 x d x-

Let I = ∫sin^6x+cos^6x/sin^2xcos^2xdx=∫(sin^2x)^3+(cos^2x)^3/sin^2x.cos^2xdx = ∫(sin^2x+cos^2x)(sin^4x sin^2xcos^2x+cos^4x)/sin^2x.cos^2xdx = ∫sin^4x/sin^2xco ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Integration by Partial Fractions

∫sin 6 x+cos ...

Question

$\int \frac{s i n^{6} x + c o s^{6} x}{s i n^{2} x c o s^{2} x} d x =$

Open in App

Solution

Let $I = \int \frac{s i n^{6} x + c o s^{6} x}{s i n^{2} x c o s^{2} x} d x = \int \frac{(s i n^{2} x)^{3} + (c o s^{2} x)^{3}}{s i n^{2} x . c o s^{2} x} d x = \int \frac{(s i n^{2} x + c o s^{2} x) (s i n^{4} x - s i n^{2} x c o s^{2} x + c o s^{4} x)}{s i n^{2} x . c o s^{2} x} d x = \int \frac{s i n^{4} x}{s i n^{2} x c o s^{2} x} d x + \int \frac{c o s^{4} x}{s i n^{2} x . c o s^{2} x} d x - \int \frac{s i n^{2} x c o s^{2} x}{s i n^{2} x . c o s^{2} x} d x = \int t a n^{2} x d x + \int c o t^{2} x d x - \int 1 d x = \int (s e c^{2} x - 1) d x + \int (c o s e c^{2} x - 1) d x - \int 1 d x = \int s e c^{2} x d x + c o s e c^{2} x d x - 3 \int d x I = t a n x - c o t x - 3 x + C$

Suggest Corrections

∫sin6x+cos6xsin2xcos2xdx=

$\int \frac{s i n^{6} x + c o s^{6} x}{s i n^{2} x c o s^{2} x} d x =$