int(d theta)/(b cos theta+c sin theta)Kaman jatakani

int (sin theta + cos theta) * d theta

int (1 + cos theta) / (theta + sin theta) d theta

int (d theta)/((sin theta - 2 cos theta)(2 sin theta + cos theta))

int (d theta)/((sin theta - 2 cos theta)(2 sin theta + cos theta))

int(d theta)/(sin theta cos^(3)theta) is equal to

Evaluate : int (2 sin 2 theta- cos theta)/(6 - cos^(2) theta- 4 sin theta) d theta

If int (sin 3 theta + sin theta ) cos theta e ^(sin theta ) d theta = (A sin ^(3) theta + B cos ^(2) theta + C sin theta + D cos theta +E)e^(sin theta)+F, then:

int(d theta)/((sin theta-2cos theta)(2sin theta+cos theta))=

Integration using rigonometric identities : int(sin theta)/(cos3 theta)+(sin 3 theta)/(cos9 theta)+( sin9 theta)/(cos 27theta)d theta=...+c

If sin theta=a/b cos theta=c/d then tan theta=