How do you integrate int(sinx+cosx)^2? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer Massimiliano Mar 1, 2015 The answer is: x-1/2cos2x+c. int(sinx+cosx)^2dx=int(sin^2x+2sinxcosx+cos^2x)dx= =int(1+sin2x)dx=intdx+1/2int2sin2xdx= =x-1/2cos2x+c. Answer link Related questions How do you find the integral int1/(x^2*sqrt(x^2-9))dx ? How do you find the integral intx^3/(sqrt(x^2+9))dx ? How do you find the integral intx^3*sqrt(9-x^2)dx ? How do you find the integral intx^3/(sqrt(16-x^2))dx ? How do you find the integral intsqrt(x^2-1)/xdx ? How do you find the integral intsqrt(x^2-9)/x^3dx ? How do you find the integral intx/(sqrt(x^2+x+1))dx ? How do you find the integral intdt/(sqrt(t^2-6t+13)) ? How do you find the integral intx*sqrt(1-x^4)dx ? How do you prove the integral formula intdx/(sqrt(x^2+a^2)) = ln(x+sqrt(x^2+a^2))+ C ? See all questions in Integration by Trigonometric Substitution Impact of this question 12064 views around the world You can reuse this answer Creative Commons License