Question

question 3
3.1 determine the following integrals:
3.1.1 $int (x^{3}-2)sqrt{x^{4}-8x} dx$
3.1.2 $int e^{\tan(2x)}sec^{2}(2x) dx$
3.1.3 $int ln(e^{2x - 1}) dx$

Explanation:

Step1: For 3.1.1, use substitution

Let $u = x^{4}-8x$, then $du=(4x^{3}-8)dx = 4(x^{3}-2)dx$, so $(x^{3}-2)dx=\frac{1}{4}du$.
The integral $\int(x^{3}-2)\sqrt{x^{4}-8x}dx=\frac{1}{4}\int\sqrt{u}du$.

Step2: Integrate $\frac{1}{4}\int\sqrt{u}du$

We know that $\int u^{n}du=\frac{u^{n + 1}}{n+1}+C$ ($n
eq - 1$), here $n=\frac{1}{2}$.
$\frac{1}{4}\int u^{\frac{1}{2}}du=\frac{1}{4}\times\frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C=\frac{1}{6}(x^{4}-8x)^{\frac{3}{2}}+C$.

Step3: For 3.1.2, use substitution

Let $t=\tan(2x)$, then $dt = 2\sec^{2}(2x)dx$, so $\sec^{2}(2x)dx=\frac{1}{2}dt$.
The integral $\int e^{\tan(2x)}\sec^{2}(2x)dx=\frac{1}{2}\int e^{t}dt$.

Step4: Integrate $\frac{1}{2}\int e^{t}dt$

Since $\int e^{t}dt=e^{t}+C$, then $\frac{1}{2}\int e^{t}dt=\frac{1}{2}e^{\tan(2x)}+C$.

Step5: For 3.1.3, simplify first

Since $\ln(e^{2x - 1})=2x-1$, then $\int\ln(e^{2x-1})dx=\int(2x - 1)dx$.

Step6: Integrate $\int(2x - 1)dx$

$\int(2x - 1)dx=\int 2xdx-\int 1dx=x^{2}-x + C$.

Answer:

3.1.1: $\frac{1}{6}(x^{4}-8x)^{\frac{3}{2}}+C$
3.1.2: $\frac{1}{2}e^{\tan(2x)}+C$
3.1.3: $x^{2}-x + C$