Question #66116 Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Monzur R. Apr 16, 2017 "A"(x)=1/3sin3x-1/3sqrt3cos3x+"c" Explanation: Recall that intcos(ax) dx=1/asinx and intsin(ax) dx=-1/acosx. intcos3x+sqrt3sin3x dx=intcos3x dx+sqrt3intsin3x dx= 1/3sin3x-1/3sqrt3cos3x+"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 1300 views around the world You can reuse this answer Creative Commons License