Question #89694 Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Andrea S. Apr 14, 2017 int cosxsqrt(1+cos2x)dx =(x +sinxcosx)/sqrt2 + C Explanation: Use the trigonometric identity: cos^2x = (1+cos2x)/2 so: cosxsqrt(1+cos2x) = cosx sqrt(2cos^2x) =sqrt2cos^2x = (1+cos2x)/sqrt2 and: int cosxsqrt(1+cos2x)dx = int (1+cos2x)/sqrt2dx int cosxsqrt(1+cos2x)dx = 1/sqrt2 int dx +1/(2sqrt2) int cos2xd(2x) int cosxsqrt(1+cos2x)dx = x/sqrt2 +(sin2x)/(2sqrt2) +C int cosxsqrt(1+cos2x)dx =(x +sinxcosx)/sqrt2 + C Answer link Related questions How do I evaluate the indefinite integral intsin^3(x)*cos^2(x)dx ? How do I evaluate the indefinite integral intsin^6(x)*cos^3(x)dx ? How do I evaluate the indefinite integral intcos^5(x)dx ? How do I evaluate the indefinite integral intsin^2(2t)dt ? How do I evaluate the indefinite integral int(1+cos(x))^2dx ? How do I evaluate the indefinite integral intsec^2(x)*tan(x)dx ? How do I evaluate the indefinite integral intcot^5(x)*sin^4(x)dx ? How do I evaluate the indefinite integral inttan^2(x)dx ? How do I evaluate the indefinite integral int(tan^2(x)+tan^4(x))^2dx ? How do I evaluate the indefinite integral intx*sin(x)*tan(x)dx ? See all questions in Integrals of Trigonometric Functions Impact of this question 1715 views around the world You can reuse this answer Creative Commons License