How do I find the general antiderivative of #f(x)= sin^2x + cos^2x#? Calculus Techniques of Integration Integration by Trigonometric Substitution 1 Answer GiĆ³ Mar 6, 2015 Actually #sin^2(x)+cos^2(x)=1# So your antiderivative is: #int[sin^2(x)+cos^2(x)]dx=int1dx=x+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 2254 views around the world You can reuse this answer Creative Commons License