Question

1) I=∫sec^2(πlnx)/3x dx, x>1
2) I=∫(tan(4ϴ)^-1/4 sec^2(4ϴ) dϴ
3) I=∫cos(2+ln√x)/x dx, x>1
4) ∫(1+√y)^1/4 /√y dy

Answer 1

$$1.\ I = \intop\frac{\sec^2(\pi lnx)}{3x}dx = \intop\frac{\sec^2(\pi lnx)}{3}d(lnx)=\\ = \frac{1}{3\pi}\intop\frac{d(\pi lnx)}{\cos^2(\pi lnx)}=\frac{\tan(\pi lnx)}{3\pi} + C \\ 2.\ I = \intop\tan(4\Theta)^{-1/4}\sec^2(4\Theta)d\Theta = \\ = \frac{1}{4}\cdot\intop\tan(4\Theta)^{-1/4}\sec^2(4\Theta)d(4\Theta) = \\ = \frac{1}{4}\cdot\intop\tan(4\Theta)^{-1/4}d(\tan(4\Theta)) = \\ =\frac{\tan(4\Theta)^{3/4}}{3}+C\\ 3.\ I = \intop\cos(2+ln\sqrt x) \frac{dx}{x}=\\ = 2\intop\cos(2+ln\sqrt x) d(ln\sqrt x)=\\ =2\sin(2+ln\sqrt x)+C \\ 4.\ I = \intop\frac{(1+\sqrt y)^{1/4}}{\sqrt y} dy = 2\intop(1+\sqrt y)^{1/4} d(\sqrt y) =\\ = 2\intop(1+\sqrt y)^{1/4} d(1+\sqrt y) =\\ = \frac{8}{5}\cdot(1+\sqrt y)^{5/4}+C$$