Question #89694 Calculus Introduction to Integration Integrals of Trigonometric Functions 1 Answer Andrea S. Apr 14, 2017 ∫cosx√1+cos2xdx=x+sinxcosx√2+C Explanation: Use the trigonometric identity: cos2x=1+cos2x2 so: cosx√1+cos2x=cosx√2cos2x=√2cos2x=1+cos2x√2 and: ∫cosx√1+cos2xdx=∫1+cos2x√2dx ∫cosx√1+cos2xdx=1√2∫dx+12√2∫cos2xd(2x) ∫cosx√1+cos2xdx=x√2+sin2x2√2+C ∫cosx√1+cos2xdx=x+sinxcosx√2+C Answer link Related questions How do I evaluate the indefinite integral ∫sin3(x)⋅cos2(x)dx ? How do I evaluate the indefinite integral ∫sin6(x)⋅cos3(x)dx ? How do I evaluate the indefinite integral ∫cos5(x)dx ? How do I evaluate the indefinite integral ∫sin2(2t)dt ? How do I evaluate the indefinite integral ∫(1+cos(x))2dx ? How do I evaluate the indefinite integral ∫sec2(x)⋅tan(x)dx ? How do I evaluate the indefinite integral ∫cot5(x)⋅sin4(x)dx ? How do I evaluate the indefinite integral ∫tan2(x)dx ? How do I evaluate the indefinite integral ∫(tan2(x)+tan4(x))2dx ? How do I evaluate the indefinite integral ∫x⋅sin(x)⋅tan(x)dx ? See all questions in Integrals of Trigonometric Functions Impact of this question 1623 views around the world You can reuse this answer Creative Commons License