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Evaluate \(\int\left(5 x^2-8 x+5\right) d x\)
\(
\int\left(5 x^2-8 x+5\right) d x=\frac{5 x^3}{3}-4 x^2+5 x+C
\)
Evaluate \(\int\left(\frac{8}{x}-\frac{5}{x^2}+\frac{6}{x^3}\right) d x\)
\(
\begin{aligned}
& \int\left(\frac{8}{x}-\frac{5}{x^2}+\frac{6}{x^3}\right) d x=\int\left(\frac{8}{x}-5 x^{-2}+6 x^{-3}\right) d x \\
& =8 \operatorname{Ln}(x)-\frac{5 x^{-1}}{-1}+\frac{6 x^{-2}}{-2}=8 \operatorname{Ln}(x)+\frac{5}{x}-\frac{3}{x^2}+C
\end{aligned}
\)
Evaluate \(\int\left(\sqrt{x}+\frac{1}{3 \sqrt{x}}\right) d x\)
\(\int\left(\sqrt{x}+\frac{1}{3 \sqrt{x}}\right) d x=\int\left(x^{\frac{1}{2}}+\frac{1}{3} x^{-\frac{1}{2}}\right) d x\)
\(
=\frac{x^{\frac{3}{2}}}{\frac{3}{2}}+\frac{1}{3} \frac{x^{\frac{1}{2}}}{\frac{1}{2}}=\frac{2}{3} x^{\frac{3}{2}}+\frac{2}{3} x^{\frac{1}{2}}+C
\)
Evaluate \(\int \frac{1}{x \sqrt{x}} d x\)
\(
\int \frac{1}{x \sqrt{x}} d x=\int x^{-\frac{3}{2}} d x=-\frac{2}{\sqrt{x}}+C
\)
Evaluate \(\int d \theta\)
\(
\int d \theta=\theta+C
\)
Evaluate \(\int 7 \sin (x) d x\)
\(
\int 7 \sin (x) d x=-7 \cos (x)+C
\)
Evaluate \(\int 12 \cos (4 \theta) d \theta\)
\(
\int 12 \cos (4 \theta) d \theta=3 \sin 4 \theta+C
\)
Evaluate \(\int 4 \sin \left(\frac{x}{3}\right) d x\)
\(
\int 4 \sin \left(\frac{x}{3}\right) d x=-12 \cos \left(\frac{x}{3}\right)+C
\)
Evaluate \(\int 4 e^{-7 x} d x\)
\(
\int 4 e^{-7 x} d x=-\frac{4 e^{-7 x}}{7}+C
\)
Evaluate \(\int-5 \cos \pi x d x\)
\(
\int-5 \cos \pi x d x=-\frac{5 \sin (\pi x)}{\pi}+C
\)
Evaluate \(\int_1^4\left(5 x^2-8 x+5\right) d x\)
\(
\int_1^4\left(5 x^2-8 x+5\right) d x=\left.\left(\frac{5 x^3}{3}-4 x^2+5 x\right)\right|_1 ^4=\frac{188}{3}-\frac{8}{3}=60
\)
Evaluate \(\int_4^9\left(\sqrt{x}+\frac{1}{3 \sqrt{x}}\right) d x\)
\(
\int_4^9\left(\sqrt{x}+\frac{1}{3 \sqrt{x}}\right) d x=\left.\left(\frac{2}{3} x^{\frac{3}{2}}+\frac{2}{3} x^{\frac{1}{2}}\right)\right|_4 ^9=20-\frac{20}{3}=\frac{40}{3}=13.333
\)
Evaluate \(\int_1^4 \frac{5}{x^3} d x\)
\(
\int_1^4 \frac{5}{x^3} d x=-\left.\frac{5}{2 x^2}\right|_1 ^4=-\frac{5}{32}+\frac{5}{2}=\frac{75}{32}=2.344
\)
Evaluate \(\int_{-2}^1\left(2 t^2-1\right)^2 d t\)
\(
\int_{-2}^1\left(2 t^2-1\right)^2 d t=\left.\left(\frac{4 t^5}{5}-\frac{4 t^3}{3}+t\right)\right|_{-2} ^1=\frac{7}{15}+\frac{254}{15}=\frac{87}{5}=17.4
\)
\(
\text { Evaluate } \int_6^0(2+5 x) \mathrm{e}^{\frac{1}{3} x} d x
\)
The first step here is to pick \(u\) and \(d v\). We want to choose \(u\) and \(d v\) so that when we compute \(d u\) and \(v\) and plugging everything into the Integration by Parts formula the new integral we get is one that we can do.
With that in mind it looks like the following choices for \(u\) and \(d v\) should work for us.
\(
u=2+5 x \quad d v=\mathbf{e}^{\frac{1}{3} x} d x
\)
Hide Step \(2-\)
Next, we need to compute \(d u\) (by differentiating \(u\) ) and \(v\) (by integrating \(d v\) ).
\(
\begin{aligned}
u & =2+5 x & & \rightarrow & d u & =5 d x \\
d v & =\mathbf{e}^{\frac{1}{3} x} d x & & \rightarrow & v & =3 \mathbf{e}^{\frac{1}{3} x}
\end{aligned}
\)
Hide Step \(3-\)
We can deal with the limits as we do the integral or we can just do the indefinite integral and then take care of the limits in the last step. We will be using the later way of dealing with the limits for this problem.
So, plugging \(u, d u, v\) and \(d v\) into the Integration by Parts formula gives,
\(
\int(2+5 x) \mathbf{e}^{\frac{1}{3} x}=(2+5 x)\left(3 \mathbf{e}^{\frac{1}{3} x}\right)-\int 5\left(3 \mathbf{e}^{\frac{1}{3} x}\right) d x=3 \mathbf{e}^{\frac{1}{3} x}(2+5 x)-15 \int \mathbf{e}^{\frac{1}{3} x} d x
\)
\(
\int(2+5 x) \mathbf{e}^{\frac{1}{3} x}=3 \mathbf{e}^{\frac{1}{3} x}(2+5 x)-45 \mathbf{e}^{\frac{1}{3} x}+c=15 x \mathbf{e}^{\frac{1}{3} x}-39 \mathbf{e}^{\frac{1}{3} x}+c
\)
The final step is then to take care of the limits.
\(
\int_6^0(2+5 x) \mathbf{e}^{\frac{1}{3} x} d x=\left.\left(15 x \mathbf{e}^{\frac{1}{3} x}-39 \mathbf{e}^{\frac{1}{3} x}\right)\right|_6 ^0=-39-51 \mathbf{e}^2=-415.8419
\)
\(\text { Evaluate } \int_0^\pi x^2 \cos (4 x) d x\)
The first step here is to pick \(u\) and \(d v\). We want to choose \(u\) and \(d v\) so that when we compute \(d u\) and \(v\) and plugging everything into the Integration by Parts formula the new integral we get is one that we can do or will at least be an integral that will be easier to deal with.
With that in mind it looks like the following choices for \(u\) and \(d v\) should work for us.
\(
u=x^2 \quad d v=\cos (4 x) d x
\)
Next, we need to compute \(d u\) (by differentiating \(u\) ) and \(v\) (by integrating \(d v\) ).
\(
\begin{aligned}
& u=x^2 \quad \rightarrow \quad d u=2 x d x \\
& d v=\cos (4 x) d x \quad \rightarrow \quad v=\frac{1}{4} \sin (4 x) \\
&
\end{aligned}
\)
So, plugging \(u, d u, v\) and \(d v\) into the Integration by Parts formula gives,
\(
\int x^2 \cos (4 x) d x=\frac{1}{4} x^2 \sin (4 x)-\frac{1}{2} \int x \sin (4 x) d x
\)
So, here are the choices for \(u\) and \(d v\) for the new integral.
\(
\begin{array}{rlrlrl}
u & =x & & \rightarrow & d u & =d x \\
d v & =\sin (4 x) d x & \rightarrow & v & =-\frac{1}{4} \cos (4 x)
\end{array}
\)
\(
\begin{aligned}
\int x^2 \cos (4 x) d x & =\frac{1}{4} x^2 \sin (4 x)-\frac{1}{2}\left[-\frac{1}{4} x \cos (4 x)+\frac{1}{4} \int \cos (4 x) d x\right] \\
& =\frac{1}{4} x^2 \sin (4 x)-\frac{1}{2}\left[-\frac{1}{4} x \cos (4 x)+\frac{1}{16} \sin (4 x)\right]+c \\
& =\frac{1}{4} x^2 \sin (4 x)+\frac{1}{8} x \cos (4 x)-\frac{1}{32} \sin (4 x)+c
\end{aligned}
\)
\(
\int_0^\pi x^2 \cos (4 x) d x=\left.\left(\frac{1}{4} x^2 \sin (4 x)+\frac{1}{8} x \cos (4 x)-\frac{1}{32} \sin (4 x)\right)\right|_0 ^\pi=\frac{1}{8} \pi
\)
Evaluate \(\int\left(1 / \sin ^2 x\right) d x\)
\(
\begin{gathered}
\int\left(1 / \sin ^2 x\right) d x=\int \operatorname{cosec}^2 x d x \\
=-\cot x+c
\end{gathered}
\)
Evaluate \(\int e^x d x\)
Evaluate \(\int\left(1+x^2\right)^{-1} d x\)
\(
\begin{gathered}
\int\left(1+x^2\right)^{-1} d x=\int 1 /\left(1+x^2\right) d x \\
=\tan ^{-1} x+c
\end{gathered}
\)
Evaluate \(\int\left(1-x^2\right)^{-1 / 2} d x\)
\(
\begin{gathered}
\int\left(1-x^2\right)^{-1 / 2} d x=\int 1 /\left(1-x^2\right)^{1 / 2} d x \\
=\int 1 / \sqrt{ }\left(1-x^2\right) d x \\
=\sin ^{-1} x+c
\end{gathered}
\)
Evaluate \(\int 1 /\left(1+36 x^2\right) d x\)
\(
\begin{aligned}
\int 1 /\left(1+36 x^2\right) d x & \\
& =\int 1 /\left(1+(6 x)^2\right) d x \\
& =\tan ^{-1}(6 x) / 6+c \\
& =(1 / 6) \tan ^{-1}(6 x)+c
\end{aligned}
\)
Evaluate \(\int 1 / \sqrt{ }\left(1-81 x^2\right) d x\)
\(
\int 1 / \sqrt{ }\left(1-81 x^2\right) d x
\)
\(
=\int 1 / \sqrt{ }\left(1-(9 x)^2\right) d x
\)
\(
=\sin ^{-1}(9 x) / 9+c
\)
\(
=(1 / 9) \sin ^{-1}(9 x)+c
\)
Integrate \(\int_0^1 \frac{1}{1+x^2} d x\)
\(
\begin{aligned}
& I=\int_0^1 \frac{1}{1+x^2} d x \\
& =\left[\tan ^{-1} x\right]_0^1 \\
& =\left[\tan ^{-1} 1-\tan ^{-1} 0\right] \\
& =\left[\frac{\pi}{4}-0\right] \\
& =\frac{\pi}{4}
\end{aligned}
\)
Integrate \(\int 2 x \cos \left(x^2-5\right) \cdot d x\)
Let, \(1=\int 2 x \cos \left(x^2-5\right) \cdot d x\)
Let \(x^2-5=t \dots(1)\)
\(
2 x \cdot d x=d t
\)
Substituting these values, we have
\(
\begin{aligned}
& I=\int \cos (t) \cdot d t \\
& =\sin t+c \ldots (2)
\end{aligned}
\)
Substituting the value of 1 in 2 , we have
\(
=\sin \left(x^2-5\right)+C
\)
Evaluate \(\int\left(x^e+e^x+e^e\right) d x\)
Let us split the above equation.
\(
\int x^e d x+\int e^x d x+\int e^e d x
\)
By the formula, we know;
\(
\int x^n d x=x^{n+1} / n+1
\)
Therefore,
\(
x^{e+1} / e+1+e^x+e^e x+c
\)
Evaluate \(\int \frac{1}{\sqrt{x}(x+1)} d x\)
This is an example of an integral that can be done by simple u-substitution, but it’s easy to miss if you’re not careful. Solve it by letting \(u=\sqrt{x}\), then \(d u=\frac{1}{\sqrt{x}}\), and \(x+1=u^2+1\). So we have
\(
2 \int \frac{d u}{u^2+1}=2 \tan ^{-1}(u)
\)
Resubstituting for u gives
\(
=2 \tan ^{-1}(\sqrt{x})+C
\)
Evaluate \(\int x \tan ^2(x) d x\)
Like a lot of integrals involving trig. functions, this one is more easily-solved by replacing the \(\tan ^2(x)\) term using the Pythagorean identity, \(\tan ^2(x)=\sec ^2(x)-1\) :
\(
\begin{aligned}
\int x \tan ^2(x) d x & =\int x\left(\sec ^2(x)-1\right) d x \\
& =\int x \sec ^2(x) d x-\int x d x
\end{aligned}
\)
Let \(u=\cos (x)\), then \(d u=-\sin (x) d x\)
\(
\begin{aligned}
\rightarrow-\int \frac{d u}{u} & =\ln (u) \\
& \rightarrow-\ln |\cos (x)|
\end{aligned}
\)
Now our full solution is
\(
=x \tan (x)+\ln |\cos (x)|-\frac{x^2}{2}+C
\)
Find \(\int e^u \sin u d u\)
Since the derivative of \(e^u\) is \(e^u\) and derivatives of sin eventually get back to sin, it is reasonable to try out integration by parts:
\(
\begin{aligned}
\int e^u \sin u d u & =-e^u \cos u+\int e^u \cos u d u \\
& =-e^u \cos u+e^u \sin u-\int e^u \sin u d u
\end{aligned}
\)
Solving for the integral we want then gives
\(
\int e^u \sin u d u=\frac{e}{2}(\sin u-\cos u)+C
\)
Evaluate \(\int \cos ^4 \varphi-\sin ^4 \varphi d \varphi\)
\(
\int \cos ^4 \varphi-\sin ^4 \varphi d \varphi
\)
Since we don’t know any nice formulas involving fourth powers of trig functions, we can try to use the Pythagorean identity \(\cos ^2 x+\sin ^2 x=1\) to reduce the powers:
\(
\begin{gathered}
\cos ^4 x=\cos ^2 x \cdot \cos ^2 x=\cos ^2 x-\cos ^2 x \sin ^2 x \\
\sin ^4 x=\sin ^2 x \cdot \sin ^2 x=\sin ^2 x-\sin ^2 x \cos ^2 x
\end{gathered}
\)
So we discover the surprising identity
\(
\cos ^4 x-\sin ^4 x=\cos ^2 x-\sin ^2 x=\cos (2 x)
\)
Our integral is then just
\(
\int \cos (2 \varphi) d \varphi=\frac{1}{2} \sin (2 \varphi)+C
\)
Evaluate \(\int \frac{\tan \theta}{\sqrt{\cos ^4 \theta-1}} d \theta\)
\(
\int \frac{\tan \theta}{\sqrt{\cos ^4 \theta-1}} d \theta
\)
As a first attempt, we might try to split up \(\tan \theta\) to see what we get:
\(
\int \frac{\tan \theta}{\sqrt{\cos ^4 \theta-1}} d \theta=\int \frac{\sin \theta}{\cos \theta \sqrt{\cos ^4 \theta-1}} d \theta
\)
This doesn’t seem much better. On the other hand, we know that things like \(d x / \sqrt{x^2-1}\) are OK to integrate, so we can try to put our integrand in that form. To do this, we would define \(x=\cos ^2 \theta\), so that \(d x=-2 \cos \theta \sin \theta d \theta\), giving
\(
\begin{aligned}
\int \frac{\sin \theta}{\cos \theta \sqrt{\cos ^4 \theta-1}} \cdot \frac{d x}{-2 \cos \theta \sin \theta} & =-\frac{1}{2} \int \frac{d x}{\cos ^2 \theta \sqrt{\cos ^4 \theta-1}} \\
& =-\frac{1}{2} \int \frac{d x}{x \sqrt{x^2-1}} \\
& =-\frac{1}{2} \sec ^{-1} x+C \\
& =\frac{-1}{2} \sec ^{-1}\left(\cos ^2 \theta\right)+C
\end{aligned}
\)
Compute \(\int \frac{x-1}{x^2-2 x+5} d x\)
By completing the square in the denominator, we will end up with terms of the form \(A x /\left(B x^2+C\right)\) or \(A /\left(B x^2+C\right)\), both of which are OK to integrate. \(x^2-2 x+5=(x-1)^2+4\), so we get (over-detailed calculation which can be simplified if you remember the \(d x / a^2+x^2\) integral)
\(
\begin{aligned}
\int \frac{x-1}{x^2-2 x+5} d x & =\int \frac{x-1}{(x-1)^2+4} d x \\
& =\frac{1}{4} \int \frac{x-1}{\left(\frac{x-1}{2}\right)^2+1} d x \\
& =\frac{1}{2} \int \frac{\frac{x-1}{2}}{\left(\frac{x-1}{2}\right)^2+1} d x
\end{aligned}
\)
Setting \(u=\left(\frac{x-1}{2}\right)^2, d u=\frac{x-1}{2} d x\), so we have
\(
\frac{1}{2} \int \frac{d u}{u^2+1}=\frac{1}{2} \tan ^{-1} u+C=\frac{1}{2} \tan ^{-1}\left(\frac{(x-1)^2}{4}\right)+C
\)
Compute \(\int \frac{\sqrt{1-\beta^2}}{\beta^2} d \beta\)
The \(\sqrt{1-\beta^2}\) makes me think of trig functions. In particular, if \(\beta=\) \(\sin \theta\), then \(\sqrt{1-\beta^2}=\cos \theta\). This is a sort of “backwards” substitution, but perfectly valid. Since \(\beta=\sin \theta, d \beta=\cos \theta d \theta\) and the integral is just
\(
\begin{aligned}
\int \frac{\sqrt{1-\beta^2}}{\beta^2} d \beta & =\int \frac{\cos \theta}{\sin ^2 \theta} \cdot \cos \theta d \theta \\
& =\int \frac{1-\sin ^2 \theta}{\sin ^2 \theta} d \theta \\
& =\int \sec ^2 \theta-1 d \theta \\
& =\tan \theta-\theta+C
\end{aligned}
\)
To get back to \(\beta\), just use the fact that since \(\beta=\sin \theta, \theta=\sin ^{-1} \beta\), so the last line is equal to \(\tan \left(\sin ^{-1} \beta\right)-\sin ^{-1} \beta+C\).
Compute \(\int x^n \log x d x\) \(\text { for } n \in\{2\}\)
Since the derivative of \(\log\) is nice and simple, we might try integration by parts with \(f=\log x\) and \(d g=x^n d x\). Starting with \(n=2\), we get
\(
\begin{aligned}
\int x^2 \log x d x & =\frac{x^3}{3} \log x-\int \frac{x^3}{3} \cdot \frac{1}{x} d x \\
& =\frac{x^3 \log x}{3}-\frac{x^3}{9}+C
\end{aligned}
\)
\(
\text { Calculate the antiderivative of } \mathrm{e}^{-\mathrm{x}}
\)
Using substitution \(t=-x\).
\(d t=-d x\) or \(d x=-d t\)
\(\int e^{-x} d x=\int-e^t d t=-e^t+C=-e^{-x}+C\)
\(
\text { Find the antiderivative of } f(x)=2 x \cos \left(x^2+1\right)
\)
Using substitution \(t=x^2+1\).
\(
\begin{aligned}
& d t=2 x d x \\
& \int 2 x \cos \left(x^2+1\right) d x=\int \cos t d t=\sin t+C \\
& =\sin \left(x^2+1\right)+C
\end{aligned}
\)
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