Question

Please help me with the below question.

Answer 1

a) Substitute
`y=x^9` and
`dy=9x^8\,dx` :

`\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}`

b) Integrate by parts:

`\displaystyle \int u\,dv = uv - \int v \, du`

Take
`u = \ln(x)` and
`dv=(dx)/(x^7)`, so that
`du=\frac{dx}x` and
`v=-\frac1{6x^6}` :

`\displaystyle \int (\ln(x))/(x^7) \, dx = -(\ln(x))/(6x^6) + \frac16 \int (dx)/(x^7) \\\\ = -(\ln(x))/(6x^6) + \frac1{36x^6} + C \\\\ = \boxed{-(6\ln(x) + 1)/(36x^6) + C}`

c) Substitute
`y=√(x+1)`, so that
`x = y^2-1` and
`dx=2y\,dy` :

`\displaystyle \frac12 \int e^(√(x+1)) \, dx = \frac12 \int 2y e^y \, dy = \int y e^y \, dy`

Integrate by parts with
`u=y` and
`dv=e^y\,dy`, so
`du=dy` and
`v=e^y` :

`\displaystyle \int ye^y \, dy = ye^y - \int e^y \, dy = ye^y - e^y + C = (y-1)e^y + C`

Then

`\displaystyle \frac12 \int e^(√(x+1)) \, dx = \boxed{\left(√(x+1)-1\right) e^(√(x+1)) + C}`

d) Integrate by parts with
`u=\sin(\pi x)` and
`dv=e^x\,dx`, so
`du=\pi\cos(\pi x)\,dx` and
`v=e^x` :

`\displaystyle \int \sin(\pi x) \, e^x \, dx = \sin(\pi x) \, e^x - \pi \int \cos(\pi x) \, e^x \, dx`

By the fundamental theorem of calculus,

`\displaystyle \int_0^1 \sin(\pi x) \, e^x \, dx = - \pi \int_0^1 \cos(\pi x) \, e^x \, dx`

Integrate by parts again, this time with
`u=\cos(\pi x)` and
`dv=e^x\,dx`, so
`du=-\pi\sin(\pi x)\,dx` and
`v=e^x` :

`\displaystyle \int \cos(\pi x) \, e^x \, dx = \cos(\pi x) \, e^x + \pi \int \sin(\pi x) \, e^x \, dx`

By the FTC,

`\displaystyle \int_0^1 \cos(\pi x) \, e^x \, dx = e\cos(\pi) - 1 + \pi \int_0^1 \sin(\pi x) \, e^x \, dx`

Then

`\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{(\pi (1+e))/(1 + \pi^2)}`

e) Expand the integrand as

`(x^2)/(x+1) = ((x^2 + 2x + 1) - (2x+1))/(x+1) = ((x+1)^2 - 2 (x+1)  + 1)/(x+1) \\\\ = x - 1 + \frac1{x+1}`

Then by the FTC,

`\displaystyle \int_0^1 (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}`

f) Substitute
`e^(7x) = \tan(y)`, so
`7e^(7x) \, dx = \sec^2(y) \, dy` :

`\displaystyle \int (e^(7x))/(e^(14x) + 1) \, dx = \frac17 \int (\sec^2(y))/(\tan^2(y) + 1) \, dy \\\\ = \frac17 \int (\sec^2(y))/(\sec^2(y)) \, dy \\\\ = \frac17 \int dy \\\\ = \frac y7 + C \\\\ = \boxed{\frac17 \tan^(-1)\left(e^(7x)\right) + C}`