2. Techniques of Integration

The integral can be written as $$\int (\cos (x){)}^3\sin (x)dx$$ In this case, we notice that if we put $u=\cos (x)$ then $\frac{du}{dx}=-\sin (x)$. Therefore, $\sin (x)dx=-du$ and then by substitution $$\begin{array}{rl}\int (\cos (x){)}^3\sin (x)dx& =\int -u^{3}du\\ & =-\frac{1}{4}{u}^4+C\\ & =-\frac{1}{4}(\cos (x){)}^4+C\end{array}$$

In this case put $u=1+x^2$ and then $\frac{du}{dx}=2x$ and thus $xdx=\frac{1}{2}du$. Now because we are changing variables from $x$ to $u$ and we are given a definite integral, we must change the limits of integration. When $x=1$ then $u=1+(1{)}^2=2$ and when $x=2$ then $u=1+2^2=5$. Therefore, $$\begin{array}{rl}{\int }_1^{2}3x\sqrt{1+x^2}dx& ={\int }_2^5\frac{3}{2}\sqrt{u}du\\ & =\frac{3}{2}\cdot \frac{2}{3}{u}^{3/2}{|}_2^5\\ & =5^{3/2}-2^{3/2}\end{array}$$

Let $$ Then $$

Choose $$ and $$.

When doing integration by parts with an indefinite integral: $$ Let $$ Then $$

Let $$ Then $$

In this example we will have to do integration by parts twice. Let $$ Then $$ To do the second integral, let $$ Then $$ and solving for $$ yields $$ or $$

Factor the odd power leaving one $$ and use identity $$: $$ Then $$ Then use substitution $$ and $$ $$

Factor the odd power leaving one $$ and use identity $$: $$ and then use substitution $$ and $$; we must change limits of integration: $$

Use half-angle formula: $$

Use the half-angle formulas: $$

Factor even power of $$ leaving one $$ and using identity $$ for the other; then use substitution with $$: $$

Factor out one $$ and one $$ term; use identity $$; then use substitution $$: $$

Compute $$

Let $$ and then $$ and from $$ the triangle is

Now make the substitutions: $$ and evaluate $$

Let $$ and then $$, and from the relationship $$ the triangle is:

Now make the substitutions $$ and evaluate: $$ Using the triangle we have $$ and $$. Therefore, $$

Let $$, then $$, and

Then $$ From the triangle: $$ and from $$ we obtain $$. Therefore, $$

Notice the $$ in front of $$. We can factor it out: $$ Then let $$ and then $$ Using $$, the triangle is

Then $$

Let $$ and the triangle is

Then $$

Let $$ and then $$, and the triangle is

Then $$

It is easier to use the substitution $$ then to use trigonometric substitution.

The denominator factors as $$ We then seek the decomposition: $$ where $$ and $$ are to be determined. Now $$ Now we equate numerators $$ equating coefficients yield the equations: $$ Solving for $$ and $$ we obtain $$ and $$. Therefore, the partial fraction decomposition is $$ and then $$ There is an alternative and quick way to find $$ and $$. From the relationship $$ and equating the numerators we obtain $$ Evaluating at $$ yields $$ and thus $$, and evaluating at $$ yields $$ and thus $$.

Write $$ and then $$ Equating the numerators we obtain $$ Evaluating both sides at $$ yields $$ and thus $$, evaluating at $$ yields $$ and thus $$, and evaluating at $$ yields $$ and thus $$. Thus $$

We seek the decomposition $$ and equating numerators yields $$ Evaluating at $$ yields $$ and thus $$ and evaluating at $$ yields $$ and thus $$. To obtain another equation we could evaluate at another convenient value of $$, say at $$: $$ and solving for $$ yields $$. Therefore, $$

The partial fraction decomposition takes the form $$

The partial fraction decomposition takes the form $$ and equating numerators we obtain $$ Evaluating both sides at $$ yields $$ and thus $$. Evaluating at $$ and $$ yield $$ and thus $$ and solving for $$ yields $$. Therefore, $$

The rational $$ is improper and thus we must use long division which yields $$ Therefore, $$

After performing long division one obtains that $$ Therefore, $$ Use partial fraction decomposition we get $$ Hence $$

We have $$. The end-points of the sub-intervals are Hence, the intervals are $$, $$, $$, and $$. Using right-end points for each sub-interval, the Riemann sum is $$ Therefore, $$

The form of the trapezoidal rule takes the form $$ First compute $$ Now compute the grid points $$ and $$: $$ Then $$

With $$, Simpson's rule takes the form $$ In this case $$ The grid points and the values of $$ at these points are: $$ Therefore, $$

We first consider $$ Then $$ Therefore, since the value of the integral is finite we say that the improper integral converges.

We first consider $$ Therefore, $$ Therefore, the improper integral converges to $$.

We compute $$ Hence, the improper integral diverges.

We need to break up the interval of integration first (choose $$ as the break-up point, any can be chosen): $$ We do each one separately and if even one diverges then the whole integral diverges. We first compute the anti-derivative using the substitution $$, $$: $$ Then $$ Now do the other integral: $$ Thus, one half of the improper integral diverges and thus the entire improper integral diverges.

It converges to $$.

The integrand has a vertical asymptote at $$. We first compute $$ and then we compute the one-sided limit: $$ Thus the improper integral diverges.

In this case, the integrand has a singularity at the point $$ which is not a boundary point of the interval $$. We therefore need to break-up the integral as follows: $$ and then we do each one separately. We compute $$ Thus, one-half of the improper integral diverges and thus the entire improper integral diverges.