What is #f(x) = int xsinx^2 + secx dx# if #f(pi/12)=-8 #?

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Answer 1 · 2017-08-14T22:11:44

The function is #f(x) = -1/2cos(2x) + ln|secx + tanx| - 8.62# approximately.

Start by separating the integrals.

#f(x) = int xsin(x^2) dx + int secx dx#

To solve the first integral, we let #u = x^2#. Then it follows that #du = 2xdx# and #dx = (du)/(2x)#.

#intxsin(x^2) = intxsinu(du)/(2x) = int1/2sinu du#

This integral is easily obtained.

#int 1/2sinu du = -1/2cosu + C = -1/2cos(x^2) + C#

Now to the second integral. This has already been derived by various sources as it is commonly encountered. The exact proof can be found here.

The integral is #intsecx dx = ln|secx + tanx| + C#

Putting all of this together, we obtain that

#f(x) = -1/2cos(x^2) + ln|secx + tanx| + C#

All that is left to do is to find the value of #C#. Using a calculator, you will find that #C ~~-8.62#.

Hopefully this helps!