Find -sin2x-cos2xsin2xcos2xdx-

I=∫sin^2x cos^2xsin^2xcos^2xdx I=∫sin^2xsin^2xcos^2xdx ∫cos^2xsin^2xcos^2xdx I=∫^2xdx ∫ cosec^2xdx I=tan x + x +C ...

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Byju's Answer

Standard XII

Mathematics

Range of Trigonometric Expressions

Find ∫sin2x...

Question

Find $\int \frac{s i n^{2} x - c o s^{2} x}{s i n^{2} x c o s^{2} x} d x$ .

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Solution

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

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

$I = \int {sec}^{2} x d x - \int c o s e c^{2} x d x$

$I = tan x + cot x + C$

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Similar questions

Q. Evaluate

\int \frac{{cos}^{2} x - {sin}^{2} x}{{sin}^{2} x {cos}^{2} x} d x

Q. Find

B

. If

\int \frac{s i n^{2} x - c o s^{2} x}{s i n^{2} x c o s^{2} x} d x = tan x + B + c

\int \frac{{sin}^{2} x - {cos}^{2} x}{{sin}^{2} x {cos}^{2} x} d x

is equal to

Choose the correct answer. $\int \frac{s i n^{2} x - c o s^{2} x}{s i n^{2} x c o s^{2} x} d x$ is equal to
$(a)tan x+cot x+C (b)tan x+cosec x+C (c)-tan x+cot x+C (d)tan x+sec x+C$

Q. Solve

\int \frac{cos 2 x}{{sin}^{2} x {cos}^{2} x} d x

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Find ∫sin2x−cos2xsin2xcos2xdx.

I=∫sin2x−cos2xsin2xcos2xdxI=∫sin2xsin2xcos2xdx−∫cos2xsin2xcos2xdxI=∫sec2xdx−∫cosec2xdxI=tanx+cotx+C

Find $\int \frac{s i n^{2} x - c o s^{2} x}{s i n^{2} x c o s^{2} x} d x$ .

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

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

$I = \int {sec}^{2} x d x - \int c o s e c^{2} x d x$

$I = tan x + cot x + C$