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How do you find the antiderivative of #int xsec^2(x^2)tan(x^2)dx# from #[0,sqrtpi/2]#?

Answer 1

#1/4#

First, let #u=x^2#. This implies that #du=2xdx#. When making this substitution, remember to plug the current into #u=x^2#. The bound of #0# stays #0# and #sqrtpi/2# becomes #(sqrtpi/2)^2=pi/4#.

Thus:

#int_0^(sqrtpi/2)xsec^2(x^2)tan(x^2)dx=1/2int_0^(pi/4)sec^2(u)tan(u)du#

Here notice that the derivative of tangent is present alongside the tangent function. Let #v=tan(u)# so #dv=sec^2(u)du#. The bounds become #0rarrtan(0)=0# and #pi/4rarrtan(pi/4)=1#.

#=1/2int_0^1vdv=1/2[v^2/2]_0^1=1/2(1^2/2-0^2/2)=1/2(1/2)=1/4#

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Answer 2

Evelyn Agard

To find the antiderivative of ( \int x \sec^2(x^2) \tan(x^2) , dx ) from ( 0 ) to ( \frac{\sqrt{\pi}}{2} ), you can use substitution. Let ( u = x^2 ), then ( du = 2x , dx ). The integral becomes ( \frac{1}{2} \int \sec^2(u) \tan(u) , du ). This integral can be evaluated as ( \frac{1}{2} \sec^2(u) + C ), where ( C ) is the constant of integration. After substituting back for ( u ) and applying the limits, you'll get the final result.

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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.