Use the substitution formula to evaluate the integral. $$ \int_{\pi / 4}^{3 \pi / 4}(4+\cot heta) \csc ^{2} heta \mathrm{~d} heta $$ $$ \frac{8}{4} $$ $$ \frac{16}{4} $$

Question

Use the substitution formula to evaluate the integral. $$ \int_{\pi / 4}^{3 \pi / 4}(4+\cot 	heta) \csc ^{2} 	heta \mathrm{~d} 	heta $$ $$ \frac{8}{4} $$ $$ \frac{16}{4} $$

UpStudy AI · Accepted Answer

We wish to evaluate
$$
\int_{\pi/4}^{3\pi/4} (4+\cot \theta) \csc^2 \theta\,d\theta.
$$
A convenient substitution is to let 
$$
u=\cot\theta.
$$
Then,
$$
\frac{du}{d\theta}=-\csc^2 \theta \quad\Longrightarrow\quad du=-\csc^2 \theta\,d\theta,
$$
so that
$$
\csc^2 \theta\,d\theta=-du.
$$

Changing the limits of integration we have:
$$
u(\pi/4)=\cot(\pi/4)=1,\qquad u(3\pi/4)=\cot(3\pi/4)=-1.
$$

Thus the integral becomes
$$
\int_{\pi/4}^{3\pi/4}(4+\cot\theta)\csc^2\theta\,d\theta
=\int_{u=1}^{u=-1}(4+u)(-du).
$$
We can factor out the negative sign:
$$
= -\int_{1}^{-1}(4+u)\,du.
$$
Reversing the limits of integration changes the sign:
$$
=-\left[-\int_{-1}^{1}(4+u)\,du\right]
=\int_{-1}^{1}(4+u)\,du.
$$

Now, we split the integral:
$$
\int_{-1}^{1}(4+u)\,du = 4\int_{-1}^{1}du + \int_{-1}^{1} u\,du.
$$
Evaluate each term:
1. For the first term,
   $$
   4\int_{-1}^{1}du = 4\Big[u\Big]_{-1}^{1} = 4(1-(-1))=4(2)=8.
   $$
2. For the second term, note that $$ u $$ is an odd function and the integration from $$-1$$ to $$1$$ gives zero:
   $$
   \int_{-1}^{1} u\,du = 0.
   $$

Thus, the value of the integral is
$$
8+0=8.
$$

Therefore, the evaluated integral is
$$
\boxed{8}.
$$
The integral evaluates to 8.

Use the substitution formula to evaluate the integral.

Upstudy AI Solution

Answer

Solution

The Deep Dive

Related Questions

Latest Calculus Questions