How do you integrate ∫cos2xdx by integration by parts.
Step- 1: Assume the variables:
The given integration is,
∫cos2xdx=∫cosx.cosxdx
Let u=cosx and v=cosx.
Step- 2: Calculate using integration by parts:
Integration by parts is given as,
∫u.vdx=u∫vdx-∫u'∫vdxdx
∴I=cosxsinx-∫-sinxsinxdx
=sinxcosx+∫sin2xdx [∵∫cosxdx=sinx;ddxcosx=-sinx]
=sinxcosx+∫1-cos2xdx [∵sin2x=1-cos2x]
=sinxcosx+∫dx-∫cos2xdx
⇒ I=sinxcosx+x-I [∵I=∫cos2xdx]
⇒ 2I=sinxcosx+x
⇒ I=sinxcosx2+x2+c[cbetheintegrationconstant]
Hence, the required answer is sinxcosx2+x2+c.
In some of the cases we can split the integrand into the sum of the two functions such that the integration of one of them by parts produces an integral which cancels the other integral. Suppose we have an integral of the type ∫[f(x)h(x)+g(x)]dx Let ∫f(x)h(x)dx=I1 and ∫g(x)dx=I2 Integrating I1 by parts, we get I1=f(x)∫h(x)dx−∫{f′(x)∫h(x)dx}dx ∫xex(1+x)2dx is equal to
In some of the cases we can split the integrand into the sum of the two functions such that the integration of one of them by parts produces an integral which cancels the other integral. Suppose we have an integral of the type ∫[f(x)h(x)+g(x)]dx Let ∫f(x)h(x)dx=I1 and ∫g(x)dx=I2 Integrating I1 by parts, we get I1=f(x)∫h(x)dx−∫{f′(x)∫h(x)dx}dx Now find ∫(1log x−1(log x)2)dx (x > 0)