a) Substitute [tex]y=x^9[/tex] and [tex]dy=9x^8\,dx[/tex] :
[tex]\displaystyle \int x^8 \cos(x^9) \, dx = \frac19 \int 9x^8 \cos(x^9) \, dx \\\\ = \frac19 \int \cos(y) \, dy \\\\ = \frac19 \sin(y) + C \\\\ = \boxed{\frac19 \sin(x^9) + C}[/tex]
b) Integrate by parts:
[tex]\displaystyle \int u\,dv = uv - \int v \, du[/tex]
Take [tex]u = \ln(x)[/tex] and [tex]dv=\frac{dx}{x^7}[/tex], so that [tex]du=\frac{dx}x[/tex] and [tex]v=-\frac1{6x^6}[/tex] :
[tex]\displaystyle \int \frac{\ln(x)}{x^7} \, dx = -\frac{\ln(x)}{6x^6} + \frac16 \int \frac{dx}{x^7} \\\\ = -\frac{\ln(x)}{6x^6} + \frac1{36x^6} + C \\\\ = \boxed{-\frac{6\ln(x) + 1}{36x^6} + C}[/tex]
c) Substitute [tex]y=\sqrt{x+1}[/tex], so that [tex]x = y^2-1[/tex] and [tex]dx=2y\,dy[/tex] :
[tex]\displaystyle \frac12 \int e^{\sqrt{x+1}} \, dx = \frac12 \int 2y e^y \, dy = \int y e^y \, dy[/tex]
Integrate by parts with [tex]u=y[/tex] and [tex]dv=e^y\,dy[/tex], so [tex]du=dy[/tex] and [tex]v=e^y[/tex] :
[tex]\displaystyle \int ye^y \, dy = ye^y - \int e^y \, dy = ye^y - e^y + C = (y-1)e^y + C[/tex]
Then
[tex]\displaystyle \frac12 \int e^{\sqrt{x+1}} \, dx = \boxed{\left(\sqrt{x+1}-1\right) e^{\sqrt{x+1}} + C}[/tex]
d) Integrate by parts with [tex]u=\sin(\pi x)[/tex] and [tex]dv=e^x\,dx[/tex], so [tex]du=\pi\cos(\pi x)\,dx[/tex] and [tex]v=e^x[/tex] :
[tex]\displaystyle \int \sin(\pi x) \, e^x \, dx = \sin(\pi x) \, e^x - \pi \int \cos(\pi x) \, e^x \, dx[/tex]
By the fundamental theorem of calculus,
[tex]\displaystyle \int_0^1 \sin(\pi x) \, e^x \, dx = - \pi \int_0^1 \cos(\pi x) \, e^x \, dx[/tex]
Integrate by parts again, this time with [tex]u=\cos(\pi x)[/tex] and [tex]dv=e^x\,dx[/tex], so [tex]du=-\pi\sin(\pi x)\,dx[/tex] and [tex]v=e^x[/tex] :
[tex]\displaystyle \int \cos(\pi x) \, e^x \, dx = \cos(\pi x) \, e^x + \pi \int \sin(\pi x) \, e^x \, dx[/tex]
By the FTC,
[tex]\displaystyle \int_0^1 \cos(\pi x) \, e^x \, dx = e\cos(\pi) - 1 + \pi \int_0^1 \sin(\pi x) \, e^x \, dx[/tex]
Then
[tex]\displaystyle \int_0^1 \sin(\pi x) \, e^x \, dx = -\pi \left(-e - 1 + \pi \int_0^1 \sin(\pi x) \, e^x \, dx\right) \\\\ \implies (1+\pi^2) \int_0^1 \sin(\pi x) \, e^x \, dx = 1 + e \\\\ \implies \int_0^1 \sin(\pi x) \, e^x \, dx = \boxed{\frac{\pi (1+e)}{1 + \pi^2}}[/tex]
e) Expand the integrand as
[tex]\dfrac{x^2}{x+1} = \dfrac{(x^2 + 2x + 1) - (2x+1)}{x+1} = \dfrac{(x+1)^2 - 2 (x+1) + 1}{x+1} \\\\ = x - 1 + \dfrac1{x+1}[/tex]
Then by the FTC,
[tex]\displaystyle \int_0^1 \frac{x^2}{x+1} \, dx = \int_0^1 \left(x - 1 + \frac1{x+1}\right) \, dx \\\\ = \left(\frac{x^2}2 - x + \ln|x+1|\right)\bigg|_0^1 \\\\ = \left(\frac12-1+\ln(2)\right) - (0-0+\ln(1)) = \boxed{\ln(2) - \frac12}[/tex]
f) Substitute [tex]e^{7x} = \tan(y)[/tex], so [tex]7e^{7x} \, dx = \sec^2(y) \, dy[/tex] :
[tex]\displaystyle \int \frac{e^{7x}}{e^{14x} + 1} \, dx = \frac17 \int \frac{\sec^2(y)}{\tan^2(y) + 1} \, dy \\\\ = \frac17 \int \frac{\sec^2(y)}{\sec^2(y)} \, dy \\\\ = \frac17 \int dy \\\\ = \frac y7 + C \\\\ = \boxed{\frac17 \tan^{-1}\left(e^{7x}\right) + C}[/tex]
