Question

Evaluate:
${\int }_0^{\pi /4}(x^3+x\cos x+\tan x+1)dx$

Answer 1

$\int xcosx=x\int cosxdx-\int [\frac{d\left(x\right)}{dx}.\int cosx.dx.]dx$ ........ (Integration by parts)

$=xsinx.-\int 1.sinxdx$

= x sin x + cos x _____________ (1)

$={\int }_0^{\frac{\pi }{4}}(x^3+xcosx+tanx+1)dx$

$={\int }_0^{\frac{\pi }{4}}{x}^3+{\int }_0^{\frac{\pi }{4}}xcosxdx+{\int }_0^{\frac{\pi }{4}}tanx.dx+{\int }_0^{\frac{\pi }{4}}1dx$

$=(\frac{x^4}{4}{)}_0^{\frac{\pi }{4}}+(xsinx+cosx{)}_0^{\frac{\pi }{4}}+(log\left|secx\right|{)}_0^{\frac{\pi }{4}}+(x{)}_0^{\frac{\pi }{4}}$

$=\frac{x^4}{4^5}+[\frac{\pi }{4}sin\frac{\pi }{4}+cos\frac{\pi }{4}-cos]+log\left|sec\frac{\pi }{4}\right|-log\left(sec\right(0\left)\right)+\frac{\pi }{4}-0$

$=\frac{x^4}{1024}+(\frac{\pi }{4}\times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-1)+log\left(\sqrt{2}\right)-0+\frac{\pi }{4}$ $(\because log1=0)$

$=\frac{x^4}{1024}+\frac{\pi }{4}(\frac{1}{\sqrt{2}}+1)+\frac{1}{\sqrt{2}}-1+\frac{1}{2}log2$