$\int (s i n^{4} x - c o s^{4} x) d x$ is equal to

- \frac{c o s 2 x}{2} + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

- \frac{s i n 2 x}{2} + c

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{s i n 2 x}{2} + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{c o s 2 x}{2} + c

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $- \frac{s i n 2 x}{2} + c$
$\int ({sin}^{4} x - {cos}^{4} x) d x$
$\int {sin}^{4} x - (1 - {sin}^{2} x)^{2} d x$
$\int {sin}^{4} x - 1 - {sin}^{4} x^{2} + 2 {sin}^{2} x d x$
$\int - (1 - 2 {sin}^{2} x) d x$
$\int - cos 2 x d x$
$= - \frac{sin 2 x}{2} + C$

Suggest Corrections

Similar questions

\int \frac{{cos}^{2} x - {sin}^{2} x}{\sqrt{1 + cos 4 x}} d x

\int \frac{cos 2 x + sin 2 x}{{(2 cos x - sin x)}^{2}} d x = \frac{- A}{1725} x - \frac{2}{5} log | 2 cos x - sin x | + 2 \frac{1}{2 - tan x} + C

\frac{cos 4 x + cos 3 x + cos 2 x}{sin 4 x + sin 3 x + sin 2 x}

lim x \to \infty \frac{{sin}^{4} x - {sin}^{2} x + 1}{{cos}^{4} x - {cos}^{2} x + 1}

lim x \to \infty \frac{{sin}^{4} x - {sin}^{2} x + 1}{{cos}^{4} x - {cos}^{2} x + 1}

Join BYJU'S Learning Program