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What is #int(sinx)/(cos^2 x) dx#?

Answer 1

Evelyn Agard

#secx+C#

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

using u-substitution: let #u=cosx#. then #du=-sinxdx# or #-du=sinxdx#

substituting into the original integral: #int-1/u^2du#

integrate that with power rule:

#1/u+C#

substitute #u=cosx#:

#1/cosx+C=secx+C#

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Answer 2

Emily Ammerman

#secx+C#

Based on the "inverse" nature of derivatives and antiderivatives (found through integrals), if we know that #d/dxsecx=secxtanx#, then #intsecxtanxdx=secx+C#.

Here, using #1/cosx=secx# and #tanx=sinx/cosx#, we see that

#intsinx/cos^2xdx=int1/cosx(sinx/cosx)dx=intsecxtanxdx=secx+C#

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Answer 3

Christopher Agresta

The integral of ( \frac{\sin(x)}{\cos^2(x)} ) with respect to ( x ) is ( -\frac{1}{\cos(x)} + C ), where ( C ) is the constant of integration.

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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.