Home > Calculus > Integrals of Trigonometric Functions

How do you integrate #int sqrttanxsec^2xdx#?

Answer 1

Emilia Aguilar

Let #u = tan(x)#, then #du = sec^2(x)dx#. Please see the explanation.

Let #u = tan(x)#, then #du = sec^2(x)dx#

#intsqrt(tan(t))sec^2dx = int(u^(1/2))du#

Integrate:

#intsqrt(tan(t))sec^2dx = 2/3u^(3/2) + C#

Reverse the substitution:

#intsqrt(tan(t))sec^2dx = 2/3(tan(x))^(3/2) + C#

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

Answer 2

Adam Adkins

To integrate ( \int \sqrt{\tan(x)} \sec^2(x) , dx ), you can use the substitution method:

Let ( u = \sqrt{\tan(x)} ), then ( du = \frac{\sec^2(x)}{2\sqrt{\tan(x)}} , dx ).

Now, substitute ( u ) and ( du ) into the integral:

[ \int \sqrt{\tan(x)} \sec^2(x) , dx = \int u \cdot 2u , du ]

[ = 2 \int u^2 , du ]

[ = \frac{2}{3} u^3 + C ]

[ = \frac{2}{3} (\sqrt{\tan(x)})^3 + C ]

Where ( C ) is the constant of integration.

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.