Home > Calculus > Definite and indefinite integrals

How to find integral of #sinx/cos^2x#?

Answer 1

Luke Alvarez

#intsinx/(cos^2x)dx=secx#

To find #intsinx/(cos^2x)dx#, let #t=cosx#, then as #dt=-sinxdx#

#intsinx/(cos^2x)dx=int(-dt)/t^2#

= #-(t^(-1))/(-1)#

= #1/t#

= #1/cosx#

= #secx#

Observe that #d/(dx)secx=secxtanx=sinx/cos^2x#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Christopher Agresta

To find the integral of sin(x)/cos^2(x), you can use the substitution method. Let u = cos(x), then du = -sin(x) dx. Substituting these into the integral, you get ∫(sin(x)/cos^2(x)) dx = ∫(-1/u^2) du. Now integrate with respect to u. The result is ∫(-1/u^2) du = 1/u + C = 1/cos(x) + C, where C is the constant of integration. Therefore, the integral of sin(x)/cos^2(x) is 1/cos(x) + C.

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.