9709. P3. Further Calculus

Differentiating tan-1x

If y = tan-1x, then x=tany.

We can differentiate both sides of this equation with respect to x (using implicit differentiation on the RHS) and then use Pythagoras’ Theorem on the result to get:

\frac{d}{dx}(tan^{-1}x)=\frac{1}{x^2+1}

Worked Examples

Find \frac{dy}{dx} in terms of x for each of the following:

  • y=tan-12x;
  • y=tan-13x;
  • y=tan-1(x/2);
  • y=tan-1(x/3).

Exercise

Answers

Integration of \frac {1}{x^2+a^2}

Based on the derivative of tan-1x, we can see that:

\int \frac{1}{x^2+1}dx = tan^{-1}x+c

\int \frac{1}{x^2+a^2}dx = \frac{1}{a}tan^{-1}(\frac{x}{a})+c

If the x2 has a coefficient, it is convenient to start by “moving it outside” the integral sign, eg: \frac {1}{2x^2+3}dx = \frac{1}{2} \int \frac{1}{x^2+\frac{3}{2}}dx

Worked Example

1.) \int \frac {5}{1+x^2} dx

2.) \int \frac {1}{1+4x^2} dx

3.) Find the indefinite integral of \frac {5} {9x^2+1}

Exercise

Answers

Integration of \frac{f'(x)}{f(x)}

\int \frac {f'(x)}{f(x)} dx = ln |f(x)|+c

As with other examples, we may sometimes need to modify an expression slightly to get it into this form, for example taking a constant outside of the integral sign.

Worked Examples

1.) \int \frac{1}{y} dy

2.) \int \frac{2x}{x^2}

3.) \int \frac{cosx}{sinx}dx

4.) \int tanx dx

5.) \int \frac {4}{x-3} dx

Exercise

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Integration by Substitution

This is a very powerful method to simplify an integration.

The integral and the limitsmust be changed to the new variable.

We must change back to the original variable at the end.

Worked Examples

1.) FInd the following indefinite integrals by making the suggested substitution. Remember to give your final answer in terms of x:

  • \int 2x(x^2+1)^5 dx, u = x^2+1
  • \int x \sqrt{2x^2-5} dx, u = 2x^2-5
  • \int \frac{x}{\sqrt{x+9}}dx, u = x+9

2.) Find the area of the shaded region for each of the following graphs:

3.) Use the substitution u = sinx + 1 to find \int cosxsinx(1+sinx)^3 dx

4.) Prove that \int \frac{1}{\sqrt{1-x^2}}dx=arcsinx+c

Exercise

Answers

Using Partial Fractions in Integration

We can use partial fractions to make integration easier.

1.) Use partial fractions to find the following integrals:

  • \int \frac{x-5}{(x+1)(x-2)} dx
  • \int \frac{8x^2-19x+1}{(2x+1)(x-2)^2} dx
  • \int {\frac{2}{1-x^2}}

Exercise

Answers

Integration by Parts

If an integral is a product of something easy to integrate and something difficult to integrate, we can use integration by parts to make it easier to integrate.

The integration by parts formula is derived by rearranging the product rule.

Integration by Parts Formula

\int u \frac{dv}{dx} dx = uv - \int v \frac{du}{dx} dx

Worked Examples

1.) Find \int x cos x dx

2.) Find \int x^2 ln x dx

Sometimes we need to use the technique more than once.

3.) Find \int x^2 e^x dx

Exercise

Mixed Exercise

Answers

Mixed Exercise Answers

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