The integral of √ sin 2 x cos 2 x is

Question

The integral of √ sin 2 x cos 2 x is (A) (1/3)( sin x)3 / 2+C (B) (1/3)( sin x)1 / 2+C (C) (1/3)( sin 2 x)3 / 2+C (D) (1/3)( sin 2 x)1 / 2+C

Tardigrade Admin · Accepted Answer

∫ √ sin 2 x cos 2 x d x Let sin 2 x=t ⇒ 2 cos 2 x=(d t/d x) ⇒ d x=(d t/2 cos 2 x) ∴ ∫ √ sin 2 x cos 2 x d x=∫ √t cos 2 x (d t/2 cos 2 x)=(1/2) ∫ √t d t =(1/2) (t1 / 2+1/((1)2+1))+C=(1/3) t3 / 2+C=(1/3)( sin 2 x)3 / 2+C

Q. The integral of $sin 2 x cos 2 x$ is

$\frac{1}{3} (sin x)^{3/2} + C$

$\frac{1}{3} (sin x)^{1/2} + C$

$\frac{1}{3} (sin 2 x)^{3/2} + C$

$\frac{1}{3} (sin 2 x)^{1/2} + C$

Q. The integral of sin2x​cos2x is

31​(sinx)3/2+C

31​(sinx)1/2+C

31​(sin2x)3/2+C

31​(sin2x)1/2+C

∫sin2x​cos2xdx Let sin2x=t ⇒2cos2x=dxdt​ ⇒dx=2cos2xdt​ ∴∫sin2x​cos2xdx=∫t​cos2x2cos2xdt​=21​∫t​dt =21​(21​+1)t1/2+1​+C=31​t3/2+C=31​(sin2x)3/2+C

Q. The integral of $sin 2 x cos 2 x$ is

$\frac{1}{3} (sin x)^{3/2} + C$

$\frac{1}{3} (sin x)^{1/2} + C$

$\frac{1}{3} (sin 2 x)^{3/2} + C$

$\frac{1}{3} (sin 2 x)^{1/2} + C$