Answer to Question #182147 in Calculus for Mark

Question

Evaluate the following integrals:

1. ∫ 𝑡 𝑐𝑜𝑠2𝑡 𝑑𝑡
2. ∫ 𝑥^3 𝑙𝑛𝑥 𝑑𝑥
3. ∫ 𝑥^3 𝑒^3𝑥 𝑑𝑥
4. ∫ 𝑠𝑖𝑛(𝑙𝑛𝑥) 𝑑𝑥
5. ∫𝑠𝑖𝑛^−1 𝑥 𝑑𝑥
6. ∫ 𝑦^3 𝑒^𝑦^2 𝑑𝑦

Answer 1

1.

$$\int t cos (2t) dt$$

Integrate by parts

$${\int}\mathtt{f}\mathtt{g}' = \mathtt{f}\mathtt{g} - {\int}\mathtt{f}'\mathtt{g}\\ f=t, g'=\cos\left(2t\right)\\ f'=1\, , g=\dfrac{\sin\left(2t\right)}{2}\\$$

Now solving:

$${\displaystyle\int}\dfrac{\sin\left(2t\right)}{2}\,\mathrm{d}t\\ \text{Substitute } u=2t \rightarrow \frac{du}{dt} =2 \rightarrow dt= \dfrac{1}{2}du:\\ = \frac{1}{4}\int \sin(u)du\\ =-cos(u)$$

Plugging in solved integrals:

$$\frac{1}{4}\int \sin(u)du\\ =-\frac{cos u}{4}\\ \text{ Recall that } u=2t\\ \therefore -\frac{cos (2t)}{4}\\$$

Plug in solved integrals,

$$\dfrac{t\sin\left(2t\right)}{2}-{\displaystyle\int}\dfrac{\sin\left(2t\right)}{2}\,\mathrm{d}t\\ =\dfrac{t\sin\left(2t\right)}{2}+\dfrac{\cos\left(2t\right)}{4}$$

Thus:

$${\displaystyle\int}t\cos\left(2t\right)\,\mathrm{d}t =\dfrac{t\sin\left(2t\right)}{2}+\dfrac{\cos\left(2t\right)}{4}+C\\ =t\cos\left(t\right)\sin\left(t\right)+\dfrac{\cos^2\left(t\right)}{2}+C$$

2.

$$\int x^3 \ln x\; dx$$

$$\text{Integrate by parts:} \\\,\\ \int \mathrm{fg}^{\prime}=\mathrm{fg}-\int \mathrm{f}^{\prime}{\mathrm{g}}\\ f=\ln (x), g^{\prime}=x^{3}\\ \quad f^{\prime}=\frac{\downarrow_{\text {steps }}}{x}, \quad g=\frac{\downarrow_{\text {steps }}}{4}\\ =\frac{x^{4} \ln (x)}{4}-\int \frac{x^{3}}{4} d x\\ \text{Now solving:}\\\,\\ \int \frac{x^{3}}{4} d x\\ \text{Apply linearity:}\\\,\\ =\frac{1}{4} \int x^{3} d x\\ \text{Now solving:}\\ \int x^{3} d x\\ \text{Apply power rule:}\\ \begin{aligned} \int x^{n} d x=& \frac{x^{n+1}}{n+1} \text { with } n=3: \\ &=\frac{x^{4}}{4} \end{aligned}\\\,\\ \text{Plug in solved integrals:}\\ \frac{1}{4} \int x^{3} d x\\ =\frac{x^{4}}{16}\\\,\\ \text{Plug in solved integrals:}\\ \frac{x^{4} \ln (x)}{4}-\int \frac{x^{3}}{4} d x\\ =\frac{x^{4} \ln (x)}{4}-\frac{x^{4}}{16}\\\,\\ \text{Thus}\\ \int x^{3} \ln (x) d x=\frac{x^{4} \ln (x)}{4}-\frac{x^{4}}{16}+C\\ =\frac{x^{4}(4 \ln (x)-1)}{16}+C$$

3.

$${\displaystyle\int}x^3\mathrm{e}^{3x}\,\mathrm{d}x$$

Using integration by parts

$$\int x^{3} \mathrm{e}^{3 x} \mathrm{d} x\\\,\\ \text{Integrate by parts:} \\ \int \mathrm{fg}^{\prime}=\mathrm{fg}-\int \mathrm{f}^{\prime} \mathrm{g}\\ \mathrm{f}=x^{3}, \mathrm{g}^{\prime}=\mathrm{e}^{3 x} \\ \mathrm{f}^{\prime}=3 x^{2}, \mathrm{~g}=\frac{\mathrm{e}^{3 x}}{3} \\ =\frac{x^{3} \mathrm{e}^{3 x}}{3}-\int x^{2} \mathrm{e}^{3 x} \mathrm{d} x\\ \,\\ \text{Now solving:}\\ \int x^{2} \mathrm{e}^{3 x} \mathrm{d} x\\\,\\ Integrate by parts: \\ \int \mathrm{fg}^{\prime}=\mathrm{fg}-\int \mathrm{f}^{\prime} \mathrm{g}\\ \mathrm{f}=x^{2}, \mathrm{~g}^{\prime}=\mathrm{e}^{3 x} \\ \mathrm{f}^{\prime}=2 x, \mathrm{~g}=\frac{\mathrm{e}^{3 x}}{3} \\ =\frac{x^{2} \mathrm{e}^{3 x}}{3}-\int \frac{2 x \mathrm{e}^{3 x}}{3} \mathrm{d} x\\ \,\\ \text{Now solving:} \\ \int \frac{2 x \mathrm{e}^{3 x}}{3} \mathrm{d} x \\\,\\ \text{Apply linearity:} =\frac{2}{3} \int x \mathrm{e}^{3 x} \mathrm{~d} x\\ \text{Now solving:}\\ \int x \mathrm{e}^{3 x} \mathrm{~d} x \\\,\\ \text{Integrate by parts:} \\ \int \mathrm{fg}^{\prime}=\mathrm{fg}-\int \mathrm{f}^{\prime} \mathrm{g}\\ \mathrm{f}=x, \mathrm{g}^{\prime}=\mathrm{e}^{3 x} \\ \mathrm{f}^{\prime}=1, \mathrm{g}=\frac{\mathrm{e}^{3 x}}{3} \\ =\frac{x \mathrm{e}^{3 x}}{3}-\int \frac{\mathrm{e}^{3 x}}{3} \mathrm{d} x \\\,\\ \text{Now solving:} \\ \int \frac{\mathrm{e}^{3 x}}{3} \mathrm{d} x \\ \text { Substitute } u=3 x \longrightarrow \frac{\mathrm{d} u}{\mathrm{d} x}=3 \longrightarrow \mathrm{d} x=\frac{1}{3} \mathrm{d} u: \\ =\frac{1}{9} \int \mathrm{e}^{u} \mathrm{d} u \\\,\\ \text{Now solving:} \\ \int \mathrm{e}^{u} \mathrm{d} u \\\,\\ \text{Apply exponential rule:}\\ \int \mathrm{a}^{u} \mathrm{~d} u= \frac{\mathrm{a}^{u}}{\ln (\mathrm{a})} \text { with } \mathrm{a}=\mathrm{e}: \\ =\mathrm{e}^{u}\\\,\\ \text{Plug in solved integrals}: \\ \frac{1}{9} \int \mathrm{e}^{u} \mathrm{~d} u \\ =\frac{\mathrm{e}^{u}}{9}\\ \\ \text{Undo substitution} u=3 x : \\ =\frac{\mathrm{e}^{3 x}}{9} \\ \text{Plug in solved integrals}: \\ \frac{x \mathrm{e}^{3 x}}{3}-\int \frac{\mathrm{e}^{3 x}}{3} \mathrm{~d} x \\ =\frac{x \mathrm{e}^{3 x}}{3}-\frac{\mathrm{e}^{3 x}}{9} \\ \text{Plug in solved integrals}: \\ \frac{2}{3} \int x \mathrm{e}^{3 x} \mathrm{~d} x \\ =\frac{2 x \mathrm{e}^{3 x}}{9}-\frac{2 \mathrm{e}^{3 x}}{27} \\ \text{Plug in solved integrals}:\\ \frac{x^{2} \mathrm{e}^{3 x}}{3}-\int \frac{2 x \mathrm{e}^{3 x}}{3} \mathrm{~d} x\\ =\frac{x^{2} \mathrm{e}^{3 x}}{3}-\frac{2 x \mathrm{e}^{3 x}}{9}+\frac{2 \mathrm{e}^{3 x}}{27}\\ \text{Plug in solved integrals}:\\ \frac{x^{3} \mathrm{e}^{3 x}}{3}-\int x^{2} \mathrm{e}^{3 x} \mathrm{~d} x\\ =\frac{x^{3} \mathrm{e}^{3 x}}{3}-\frac{x^{2} \mathrm{e}^{3 x}}{3}+\frac{2 x \mathrm{e}^{3 x}}{9}-\frac{2 \mathrm{e}^{3 x}}{27}\\ \text{Thus:}\\ \int x^{3} \mathrm{e}^{3 x} \mathrm{~d} x=\\\frac{x^{3} \mathrm{e}^{3 x}}{3}-\frac{x^{2} \mathrm{e}^{3 x}}{3}+\frac{2 x \mathrm{e}^{3 x}}{9}-\frac{2 \mathrm{e}^{3 x}}{27}+C\\ =\frac{\left(9 x^{3}-9 x^{2}+6 x-2\right) \mathrm{e}^{3 x}}{27}+C$$

4.

$$\int{\sin{\left(\ln{\left(x \right)} \right)} d x}$$

$$\begin{array}{c} \text { Substitute } u=\ln (x) \longrightarrow \frac{\mathrm{d} u}{\mathrm{~d} x}=\frac{1}{x}=\mathrm{e}^{u} \longrightarrow \mathrm{d} x=x \mathrm{~d} u, \text { use: } \\ =\int \mathrm{e}^{u} \sin (u) \mathrm{d} u \end{array} \\ \text{ We will integrate by parts twice in a row: } \\ \int \mathrm{fg}^{\prime}=\mathrm{fg}-\int \mathrm{f}^{\prime} \mathrm{g}\\ \text{ First time: } \\ \mathbf{f}=\sin (u), \quad \mathrm{g}^{\prime}=\mathrm{e}^{u} \\ \begin{array}{c}\\ \mathbf{f}^{\prime}=\cos (u), \mathrm{g}=\mathrm{e}^{u} \\ =\mathrm{e}^{u} \sin (u)-\int \mathrm{e}^{u} \cos (u) \mathrm{d} u \end{array} \\ \text{ Second time:} \\ \begin{array}{c}\\ \mathrm{f}=\cos (u), \quad \mathrm{g}^{\prime}=\mathrm{e}^{u} \\ \mathrm{f}^{\prime}=-\sin (u), \mathrm{g}=\mathrm{e}^{u}: \\ =\mathrm{e}^{u} \sin (u)-\left(\mathrm{e}^{u} \cos (u)-\int-\mathrm{e}^{u} \sin (u) \mathrm{d} u\right) \end{array}\\ \text { Apply linearity: }\\ =\mathrm{e}^{u} \sin (u)-\left(\mathrm{e}^{u} \cos (u)+\int \mathrm{e}^{u} \sin (u) \mathrm{d} u\right)\\ \text{ The integral } \int \mathrm{e}^{u} \sin (u) \mathrm{d} u\\ \text{ appears again on the right side of the equation, we can solve for it: }\\ =\frac{\mathrm{e}^{u} \sin (u)-\mathrm{e}^{u} \cos (u)}{2}\\ \text{ Undo substitution } u=\ln (x), \text { use: } \\ \begin{array}{c}\mathrm{e}^{\ln (x)}=x \\ =\frac{x \sin (\ln (x))-x \cos (\ln (x))}{2}\end{array} \\ \text{Thus }\\ \int \sin (\ln (x)) \mathrm{d} x =\frac{x \sin (\ln (x))-x \cos (\ln (x))}{2}+C\\ \int \sin (\ln (x)) \mathrm{d} x=\frac{x(\sin (\ln (x))-\cos (\ln (x)))}{2}+C\\$$

5.

$$\int \arcsin (x) \mathrm{d} x\\ \text{ Integrate by parts: }\\ \int \mathrm{fg}^{\prime}=\mathrm{fg}-\int \mathrm{f}^{\prime} \mathrm{g}\\ f=\arcsin (x), g^{\prime}=1 \\ {f^{\prime}}=\frac{1}{\sqrt{1-x^{2}}}, g=x \\ =x \arcsin (x)-\int \frac{x}{\sqrt{1-x^{2}}} \mathrm{~d} x\\ \text{ Now solving: }\\ \int \frac{x}{\sqrt{1-x^{2}}} d x\\ \text { Substitute } u=1-x^{2} \longrightarrow \frac{\mathrm{d} u}{\mathrm{~d} x}=-2 x \longrightarrow \mathrm{d} x=-\frac{1}{2 x} \mathrm{~d} u \text { : } \\ =-\frac{1}{2} \int \frac{1}{\sqrt{u}} \mathrm{~d} u\\ \text{ Now solving: }\\ \int \frac{1}{\sqrt{u}} \mathrm{~d} u \text{ Apply power rule: }\\ \int u^{\mathrm{n}} \mathrm{d} u=\frac{u^{\mathrm{n}+1}}{\mathrm{n}+1} \text { with } \mathrm{n}=-\frac{1}{2}: \\ =2 \sqrt{u} \\ \text{ Plug in solved integrals: }\\ -\frac{1}{2} \int \frac{1}{\sqrt{u}} \mathrm{~d} u\\ =-\sqrt{u} \\ \text{ Undo substitution } \boldsymbol{u}=\mathbf{1}-\boldsymbol{x}^{2}\\ =-\sqrt{1-x^{2}}\\ \text{ Plug in solved integrals :}\\ x \arcsin (x)-\int \frac{x}{\sqrt{1-x^{2}}} \mathrm{~d} x\\ =x \arcsin (x)+\sqrt{1-x^{2}}\\ \text{ Thus: }\\ \int \arcsin (x) \mathrm{d} x =x \arcsin (x)+\sqrt{1-x^{2}}+C$$

6.

$$\int y^{3} \mathrm{e}^{y^{2}} \mathrm{~d} y \\ \text { Substitute } u=y^{2} \longrightarrow \frac{\mathrm{d} u}{\mathrm{~d} y}=2 y \longrightarrow \mathrm{d} y=\frac{1}{2 y} \mathrm{~d} u \text { : } \\= \frac{1}{2} \int u \mathrm{e}^{u} \mathrm{~d} u\\ \text{ Now solving: } \int u \mathrm{e}^{u} \mathrm{~d} u\\ \text{ Integrate by parts: } \\ \int \mathrm{fg}^{\prime}=\mathrm{fg}-\int \mathrm{f}^{\prime} \mathrm{g}\\ \begin{array}{l} \mathrm{f}=u, \quad \mathrm{~g}^{\prime}=\mathrm{e}^{u} \\ \mathrm{f}^{\prime}=1, \mathrm{~g}=\mathrm{e}^{u} \\ =u \mathrm{e}^{u}-\int \mathrm{e}^{u} \mathrm{~d} u \end{array}\\ \text{ Now solving: } \int \mathrm{e}^{u} \mathrm{~d} u\\ \text{ Apply exponential rule: }\\ \int a^{u} \mathrm{~d} u= \frac{\mathrm{a}^{u}}{\ln (\mathrm{a})} \text { with } \mathrm{a}=\mathrm{e}: \\ =\mathrm{e}^{u} \\ \text{ Plug in solved integrals: }\\ u \mathrm{e}^{u}-\int \mathrm{e}^{u} \mathrm{~d} u \\ =u \mathrm{e}^{u}-\mathrm{e}^{u}\\ \text{ Plug in solved integrals: }\\ \frac{1}{2} \int u \mathrm{e}^{u} \mathrm{~d} u \\ = \frac{u \mathrm{e}^{u}}{2}-\frac{\mathrm{e}^{u}}{2}\\ \text{ Undo substitution }\boldsymbol{u}=\boldsymbol{y}^{2} :\\ =\frac{y^{2} \mathrm{e}^{y^{2}}}{2}-\frac{\mathrm{e}^{y^{2}}}{2}\\ \text{ Thus: }\\ \int y^{3} \mathrm{e}^{y^{2}} \mathrm{~d} y =\frac{y^{2} \mathrm{e}^{y^{2}}}{2}-\frac{\mathrm{e}^{y^{2}}}{2}+C\\ \int y^{3} \mathrm{e}^{y^{2}} \mathrm{~d} y=\frac{\left(y^{2}-1\right) \mathrm{e}^{y^{2}}}{2}+C$$