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Question: How do you integrate \( \int {\cos ^2}x \) by integration by parts method?...

How do you integrate ∫cos⁡2x\int {\cos ^2}x by integration by parts method?

Explanation

Solution

Hint : To solve the given integral expression by parts method, first we will assume the given trigonometric term of the integral as uu and then differentiate it and again assume it in another part. And finally integrate both the assumed terms separately until it's in the simplest form.

Complete step-by-step answer :
If we really want to integrate by parts choose:
∵u=cos⁡x ⇒dudx=−sin⁡x ⇒du=−sin⁡x.dx   \because u = \cos x \\\ \Rightarrow \dfrac{{du}}{{dx}} = - \sin x \\\ \Rightarrow du = - \sin x.dx \;
Now assume,
v=sin⁡x ⇒dvdx=cos⁡x ⇒dv=cos⁡x.dx   v = \sin x \\\ \Rightarrow \dfrac{{dv}}{{dx}} = \cos x \\\ \Rightarrow dv = \cos x.dx \;
Now,
∫udv=u.v−∫v.du\int udv = u.v - \int v.du
In the above equation, if we elaborate the L.H.S, then the R.H.S is as it is.
⇒∫cos⁡x.cos⁡xdx=cos⁡x.sin⁡x−∫sin⁡x(−sin⁡x.dx) ⇒∫cos⁡2x.dx=cos⁡x.sin⁡x+∫(1−cos⁡2x)dx ⇒∫cos⁡2x.dx=cos⁡x.sin⁡x+∫1dx−∫cos⁡2x.dx   \Rightarrow \int \cos x.\cos xdx = \cos x.\sin x - \int \sin x( - \sin x.dx) \\\ \Rightarrow \int {\cos ^2}x.dx = \cos x.\sin x + \int (1 - {\cos ^2}x)dx \\\ \Rightarrow \int {\cos ^2}x.dx = \cos x.\sin x + \int 1dx - \int {\cos ^2}x.dx \;
Now for the second part of the R.H.S: integrate it separately:
2. ∫cos⁡2x.dx=cos⁡x.sin⁡x+x\int {\cos ^2}x.dx = \cos x.\sin x + x
And assume: ∫cos⁡2x.dx\int {\cos ^2}x.dx as II :
∴I=∫cos⁡2x=12x+12sin⁡x.dx\therefore I = \int {\cos ^2}x = \dfrac{1}{2}x + \dfrac{1}{2}\sin x.dx
However, a shorter way is to use the trigonometric identities:
cos⁡2x=cos⁡2x−sin⁡2x=2cos⁡2x−1=1−2sin⁡2x\cos 2x = {\cos ^2}x - {\sin ^2}x = 2{\cos ^2}x - 1 = 1 - 2{\sin ^2}x and
sin⁡2x=2sin⁡x.cos⁡x\sin 2x = 2\sin x.\cos x
Now, use the upper identities in the main equation:
∴∫cos⁡2x=∫cos⁡x.sin⁡x+x−12x−12sin⁡x.dx ⇒∫cos⁡2x.dx=∫1+cos⁡2x2dx ⇒∫cos⁡2x.dx=∫12dx+12∫cos⁡2x.dx ⇒∫cos⁡2x.dx=12x+12sin⁡2x+C ∴∫cos⁡2x.dx=x2+12sin⁡x.cos⁡x+C   \therefore \int {\cos ^2}x = \int \cos x.\sin x + x - \dfrac{1}{2}x - \dfrac{1}{2}\sin x.dx \\\ \Rightarrow \int {\cos ^2}x.dx = \int \dfrac{{1 + \cos 2x}}{2}dx \\\ \Rightarrow \int {\cos ^2}x.dx = \int \dfrac{1}{2}dx + \dfrac{1}{2}\int \cos 2x.dx \\\ \Rightarrow \int {\cos ^2}x.dx = \dfrac{1}{2}x + \dfrac{1}{2}\sin 2x + C \\\ \therefore \int {\cos ^2}x.dx = \dfrac{x}{2} + \dfrac{1}{2}\sin x.\cos x + C \;
So, the correct answer is “x2+12sin⁡x.cos⁡x+C\dfrac{x}{2} + \dfrac{1}{2}\sin x.\cos x + C”.

Note : Integration by Parts is a special method of integration that is often useful when two functions are multiplied together, but is also helpful in other ways. This method is also termed as partial integration. Another method to integrate a given function is integration by substitution method. These methods are used to make complicated integrations easy.