Find -sin2x-cos2xsin2x cos2xdx

Let I=∫sin^2x cos^2xsin^2x cos^2xdx =∫sin^2xsin^2x cos^2xdx ∫cos^2xsin^2x cos^2xdx =∫^2 x dx ∫^2x dx =tan x + x +C ...

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Byju's Answer

Standard XII

Mathematics

Theorems on Integration

Find ∫sin2x...

Question

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

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Solution

Let $I = \int \frac{{sin}^{2} x - {cos}^{2} x}{{sin}^{2} x {cos}^{2} x} d x$

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

$= \int {sec}^{2} x d x - \int {csc}^{2} x d x$

$= tan x + cot x + C$

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Similar questions

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

Q. Find

\frac{d y}{d x}

when

y = \frac{1 + sin 2 x - cos 2 x}{1 + sin 2 x + cos 2 x}

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Find ∫sin2x−cos2xsin2xcos2xdx

Let I=∫sin2x−cos2xsin2xcos2xdx=∫sin2xsin2xcos2xdx−∫cos2xsin2xcos2xdx=∫sec2x dx−∫csc2x dx=tanx+cotx+C

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

Let $I = \int \frac{{sin}^{2} x - {cos}^{2} x}{{sin}^{2} x {cos}^{2} x} d x$

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

$= \int {sec}^{2} x d x - \int {csc}^{2} x d x$

$= tan x + cot x + C$