Use a substitution. Let u = 1/2x.

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here are the solutions to the integration problems:

Integrate $\int \sin\left(\frac{1}{2}x\right)\,dx$

Use a substitution.
Let $u = \frac{1}{2}x$ .
Then, $du = \frac{1}{2}\,dx$ , which means $dx = 2\,du$ .
Substitute $u$ and $dx$ into the integral.
$$ \int \sin(u) (2\,du) = 2 \int \sin(u)\,du $$
Integrate with respect to $u$ .
$$ 2 (-\cos(u)) + C = -2\cos(u) + C $$
Substitute back $u = \frac{1}{2}x$ .
$$ \boxed{-2\cos\left(\frac{1}{2}x\right) + C} $$
2. Integrate $\int \cot x\,dx$
Rewrite $\cot x$ in terms of sine and cosine.
$$ \int \frac{\cos x}{\sin x}\,dx $$
Use a substitution.
Let $u = \sin x$ .
Then, $du = \cos x\,dx$ .
Substitute $u$ and $du$ into the integral.
$$ \int \frac{1}{u}\,du $$
Integrate with respect to $u$ .
$$ \ln|u| + C $$
Substitute back $u = \sin x$ .
$$ \boxed{\ln|\sin x| + C} $$
3. Integrate $\int x \sqrt{1 + x^2}\,dx$
Use a substitution.
Let $u = 1 + x^2$ .
Then, $du = 2x\,dx$ , which means $x\,dx = \frac{1}{2}\,du$ .

Substitute $u$ and $x\,dx$ into the integral.

$$
\int \sqrt{u} \left(\frac{1}{2}\,du\right) = \frac{1}{2} \int u^{1/2}\,du
$$

Integrate with respect to $u$ .

$$
\frac{1}{2} \left(\frac{u^{1/2 + 1}}{1/2 + 1}\right) + C = \frac{1}{2} \left(\frac{u^{3/2}}{3/2}\right) + C
$$
$$
= \frac{1}{2} \cdot \frac{2}{3} u^{3/2} + C = \frac{1}{3} u^{3/2} + C
$$

Substitute back $u = 1 + x^2$ .
$$ \boxed{\frac{1}{3}(1 + x^2)^{3/2} + C} $$
4. Integrate $\int e^x (3 + e^x)^4\,dx$
Use a substitution.
Let $u = 3 + e^x$ .
Then, $du = e^x\,dx$ .
Substitute $u$ and $du$ into the integral.
$$ \int u^4\,du $$

Integrate with respect to $u$ .

$$
\frac{u^{4+1}}{4+1} + C = \frac{u^5}{5} + C
$$

Substitute back $u = 3 + e^x$ .
$$ \boxed{\frac{1}{5}(3 + e^x)^5 + C} $$
5. Integrate $\int \frac{1}{1 + x^2}\,dx$
Recognize this as a standard integral.
The derivative of $\arctan x$ is $\frac{1}{1 + x^2}$.
Apply the integration formula directly.
$$ \boxed{\arctan x + C} $$
6. Integrate $\int \cos^2 x\,dx$
Use the trigonometric identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$ .
$$ \int \frac{1 + \cos(2x)}{2}\,dx = \frac{1}{2} \int (1 + \cos(2x))\,dx $$

Integrate term by term.

$$
\frac{1}{2} \left(\int 1\,dx + \int \cos(2x)\,dx\right)
$$

Integrate each term.
For $\int \cos(2x)\,dx$ , let $u = 2x$ , so $du = 2\,dx$ , or $dx = \frac{1}{2}\,du$ . $\int \cos(2x)\,dx = \int \cos(u) \left(\frac{1}{2}\,du\right) = \frac{1}{2} \sin(u) + C_1 = \frac{1}{2} \sin(2x) + C_1$ And $\int 1\,dx = x + C_2$ .

Combine the results.

$$
\frac{1}{2} \left(x + \frac{1}{2}\sin(2x)\right) + C
$$
$$
\boxed{\frac{1}{2}x + \frac{1}{4}\sin(2x) + C}
$$

7. Integrate $\int \cos^3 x\,dx$

Rewrite $\cos^3 x$ using the identity $\cos^2 x = 1 - \sin^2 x$ .
$$ \int \cos^2 x \cdot \cos x\,dx = \int (1 - \sin^2 x) \cos x\,dx $$
Use a substitution.
Let $u = \sin x$ .
Then, $du = \cos x\,dx$ .
Substitute $u$ and $du$ into the integral.
$$ \int (1 - u^2)\,du $$
Integrate with respect to $u$ .
$$ u - \frac{u^3}{3} + C $$
Substitute back $u = \sin x$ .
$$ \boxed{\sin x - \frac{1}{3}\sin^3 x + C} $$

Use a substitution. Let u = (1)/(2)x.

Step-by-step solution

Use a substitution.

Substitute $u$ and $dx$ into the integral.

Integrate with respect to $u$ .

Substitute back $u = \frac{1}{2}x$ .

Rewrite $\cot x$ in terms of sine and cosine.

Use a substitution.

Substitute $u$ and $du$ into the integral.

Integrate with respect to $u$ .

Substitute back $u = \sin x$ .

Use a substitution.

Substitute $u$ and $x\,dx$ into the integral.

Integrate with respect to $u$ .

Substitute back $u = 1 + x^2$ .

Use a substitution.

Substitute $u$ and $du$ into the integral.

Integrate with respect to $u$ .

Substitute back $u = 3 + e^x$ .

Recognize this as a standard integral.

Apply the integration formula directly.

Use the trigonometric identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$ .

Integrate term by term.

Integrate each term.

Combine the results.

Rewrite $\cos^3 x$ using the identity $\cos^2 x = 1 - \sin^2 x$ .

Use a substitution.

Substitute $u$ and $du$ into the integral.

Integrate with respect to $u$ .

Substitute back $u = \sin x$ .

Still stuck on this one?

More Mathematics Questions

Use a substitution. Let u = (1)/(2)x.

Step-by-step solution

Use a substitution.

Substitute uuu and dxdxdx into the integral.

Integrate with respect to uuu.

Substitute back u=12xu = \frac{1}{2}xu=21​x.

Rewrite cot⁡x\cot xcotx in terms of sine and cosine.

Use a substitution.

Substitute uuu and dududu into the integral.

Integrate with respect to uuu.

Substitute back u=sin⁡xu = \sin xu=sinx.

Use a substitution.

Substitute uuu and x dxx\,dxxdx into the integral.

Integrate with respect to uuu.

Substitute back u=1+x2u = 1 + x^2u=1+x2.

Use a substitution.

Substitute uuu and dududu into the integral.

Integrate with respect to uuu.

Substitute back u=3+exu = 3 + e^xu=3+ex.

Recognize this as a standard integral.

Apply the integration formula directly.

Use the trigonometric identity cos⁡2x=1+cos⁡(2x)2\cos^2 x = \frac{1 + \cos(2x)}{2}cos2x=21+cos(2x)​.

Integrate term by term.

Integrate each term.

Combine the results.

Rewrite cos⁡3x\cos^3 xcos3x using the identity cos⁡2x=1−sin⁡2x\cos^2 x = 1 - \sin^2 xcos2x=1−sin2x.

Use a substitution.

Substitute uuu and dududu into the integral.

Integrate with respect to uuu.

Substitute back u=sin⁡xu = \sin xu=sinx.

Still stuck on this one?

More Mathematics Questions

Substitute $u$ and $dx$ into the integral.

Integrate with respect to $u$ .

Substitute back $u = \frac{1}{2}x$ .

Rewrite $\cot x$ in terms of sine and cosine.

Substitute $u$ and $du$ into the integral.

Integrate with respect to $u$ .

Substitute back $u = \sin x$ .

Substitute $u$ and $x\,dx$ into the integral.

Integrate with respect to $u$ .

Substitute back $u = 1 + x^2$ .

Substitute $u$ and $du$ into the integral.

Integrate with respect to $u$ .

Substitute back $u = 3 + e^x$ .

Use the trigonometric identity $\cos^2 x = \frac{1 + \cos(2x)}{2}$ .

Rewrite $\cos^3 x$ using the identity $\cos^2 x = 1 - \sin^2 x$ .

Substitute $u$ and $du$ into the integral.

Integrate with respect to $u$ .

Substitute back $u = \sin x$ .