Question

Find the integrals (primitives):
$\rm \int \sin^{2}x\ dx$

Answer 1

❤Heir is your answer see in pic

Answer 2

Answer:

$\rm \int \sin^{2}x\ dx =\dfrac x2 -  \dfrac {sin2x}4 + C$

Step-by-step explanation:

$\rm \int \sin^{2}x\ dx = \rm \int (1-\cos^{2}x)\ dx\\= \int dx - \int cos^2x dx ...(1)$

By the formula of cos2x

$cos2x = 2cos^2x - 1\\cos2x + 1 = 2cos^2x\\\\\dfrac {cos2x + 1}{2} = cos^2x$

$put\ the\ value\ of\ cos^2x\ in\ equation\ (1)$

$=\int dx - \int \dfrac {cos2x + 1}{2} dx\\= x - \dfrac 12 \int cos2x -\dfrac12\int dx\\\\= x -\dfrac 12\times\dfrac{sin2x}{2}} - \dfrac x2 + C\\= x -\dfrac x2 -\dfrac {sin2x}4 + C\\= \dfrac x2 -\dfrac {sin2x}4 + C$

$\rm \int \sin^{2}x\ dx =\dfrac x2 -\dfrac {sin2x}4 + C$