Integration Questions (With detailed Solutions)

Integration Questions

Question 1

Evaluate the following integrals:

(a) \(\displaystyle \int_{0}^{1} \frac{1}{(1+x^2)^2} \, dx\)

(b) \(\displaystyle \int_{2}^{4} \frac{x+2}{\sqrt{3x-4}} \, dx\)

Question 2

(a) The region enclosed by the curve \( y = x^2 - 2x \) and \( y = 3 \) is rotated about \( y = 3 \). Find the resulting volume.

(b) Find the length of a curve given by \( x = a \cos \theta \) and \( y = a \sin \theta \) from \( \theta = 0 \) to \( \theta = 3 \).

Question 3

(a) (i) \(\displaystyle \int \frac{2 \cos x + 3 \sin x}{\cos x + \sin x} \, dx\)

(ii) \(\displaystyle \int \frac{\cos x}{x} \, dx\)

(b) Prove that \(\displaystyle \int_{1}^{2} \tan^{-1} \sqrt{x^2 - 1} \, dx = \frac{2\pi}{3} \ln (2+\sqrt{3})\)

(c) Find the volume generated when a semi-circle is revolved through \( x \)-axis.

Question 4

(a) \(\displaystyle \int \frac{dx}{1 + 3e^x + 2e^{2x}}\)

(b) Evaluate the length of an arc given by the parametric equation \( x = \sqrt{2} \cos t \) and \( y = \sqrt{2} \sin t \) at the interval of \( t = 0 \) and \( t = \frac{\pi}{2} \)

(c) Find the value of the area of the regions bounded by the graph \( y = 4x^2 \) and \( y = 2\sqrt{x} \)

Question 5

(a) By writing \( x = a \cos^2 \theta + b \sin^2 \theta \) show that \(\displaystyle \int_{a}^{b} \frac{dx}{\sqrt{x-a}\sqrt{x-b}}\)

(b) Find the area bounded between the curve \( f(x) = 2x^2 + 3x + 1 \) and the lines \( x = 2 \) and \( x = 4 \)

(c) Integrate the following:

(i) \(\displaystyle \int \frac{1}{2 + \cos y + \sin y} \, dy\)

(ii) \(\displaystyle \int \frac{\cos p + \sin p}{2 \cos p + 3 \sin p} \, dp\)

Advanced Mathematics - Integration Solutions

Question 1

(a) Evaluate \(\displaystyle \int_{0}^{1} \frac{1}{(1+x^2)^2} \, dx\)

Step 1: Use trigonometric substitution

Let \( x = \tan \theta \), then \( dx = \sec^2 \theta \, d\theta \)

Step 2: Change limits

When \( x = 0 \), \( \theta = 0 \)

When \( x = 1 \), \( \theta = \frac{\pi}{4} \)

Step 3: Substitute and simplify

\[ \int_{0}^{\pi/4} \frac{\sec^2 \theta}{(1+\tan^2 \theta)^2} d\theta = \int_{0}^{\pi/4} \frac{\sec^2 \theta}{\sec^4 \theta} d\theta = \int_{0}^{\pi/4} \cos^2 \theta \, d\theta \]

Step 4: Use trigonometric identity

\[ \cos^2 \theta = \frac{1+\cos 2\theta}{2} \]

So the integral becomes:

\[ \frac{1}{2} \int_{0}^{\pi/4} (1 + \cos 2\theta) d\theta = \frac{1}{2} \left[ \theta + \frac{\sin 2\theta}{2} \right]_0^{\pi/4} \]

Step 5: Evaluate

\[ \frac{1}{2} \left( \frac{\pi}{4} + \frac{1}{2} \right) = \frac{\pi}{8} + \frac{1}{4} \]

Final answer: \(\boxed{\dfrac{\pi}{8} + \dfrac{1}{4}}\)

(b) Evaluate \(\displaystyle \int_{2}^{4} \frac{x+2}{\sqrt{3x-4}} \, dx\)

Step 1: Substitution

Let \( u = 3x - 4 \), then \( du = 3dx \) or \( dx = \frac{du}{3} \)

Step 2: Change limits

When \( x = 2 \), \( u = 2 \)

When \( x = 4 \), \( u = 8 \)

Step 3: Rewrite numerator

\[ x + 2 = \frac{1}{3}(u + 4) + 2 = \frac{u + 10}{3} \]

Step 4: Substitute into integral

\[ \int \frac{\frac{u+10}{3}}{\sqrt{u}} \cdot \frac{du}{3} = \frac{1}{9} \int (u^{1/2} + 10u^{-1/2}) du \]

Step 5: Integrate

\[ \frac{1}{9} \left( \frac{2}{3}u^{3/2} + 20u^{1/2} \right) + C \]

Step 6: Evaluate definite integral

\[ \frac{1}{9} \left[ \left( \frac{2}{3}(8)^{3/2} + 20(8)^{1/2} \right) - \left( \frac{2}{3}(2)^{3/2} + 20(2)^{1/2} \right) \right] \]

\[ = \frac{1}{9} \left( \frac{128}{3} + 40\sqrt{2} - \frac{8}{3} - 20\sqrt{2} \right) = \frac{40}{9} + \frac{20\sqrt{2}}{9} \]

Final answer: \(\boxed{\dfrac{40}{9} + \dfrac{20\sqrt{2}}{9}}\)

Question 2

(a) Find volume of revolution about y=3

Step 1: Find intersection points

\[ x^2 - 2x = 3 \Rightarrow x = -1 \text{ and } x = 3 \]

Step 2: Set up volume integral using washer method

\[ V = \pi \int_{-1}^{3} \left[ 3^2 - (3 - (x^2 - 2x))^2 \right] dx = \pi \int_{-1}^{3} (9 - (x^2 - 2x - 3)^2) dx \]

Step 3: Expand the integrand

\[ (x^2 - 2x - 3)^2 = x^4 - 4x^3 - 2x^2 + 12x + 9 \]

So the integrand becomes:

\[ 9 - x^4 + 4x^3 + 2x^2 - 12x - 9 = -x^4 + 4x^3 + 2x^2 - 12x \]

Step 4: Integrate term by term

\[ \pi \left[ -\frac{x^5}{5} + x^4 + \frac{2x^3}{3} - 6x^2 \right]_{-1}^{3} \]

Step 5: Evaluate at bounds

After calculation, we get:

\[ V = \frac{512\pi}{15} \]

Final answer: \(\boxed{\dfrac{512\pi}{15}}\)

(b) Arc length of parametric curve

Given: \( x = a \cos \theta \), \( y = a \sin \theta \) from \( \theta = 0 \) to \( \theta = 3 \)

Step 1: Compute derivatives

\[ \frac{dx}{d\theta} = -a \sin \theta, \quad \frac{dy}{d\theta} = a \cos \theta \]

Step 2: Arc length formula

\[ L = \int_{0}^{3} \sqrt{(-a \sin \theta)^2 + (a \cos \theta)^2} \, d\theta = \int_{0}^{3} \sqrt{a^2 (\sin^2 \theta + \cos^2 \theta)} \, d\theta \]

Step 3: Simplify using trigonometric identity

\[ \sin^2 \theta + \cos^2 \theta = 1 \]

So:

\[ L = \int_{0}^{3} a \, d\theta = a(3 - 0) = 3a \]

Final answer: \(\boxed{3a}\)

Question 3

(a)(i) Evaluate \(\displaystyle \int \frac{2 \cos x + 3 \sin x}{\cos x + \sin x} \, dx\)

Step 1: Express numerator as combination of denominator and its derivative

Let \( 2\cos x + 3\sin x = A(\cos x + \sin x) + B(-\sin x + \cos x) \)

Step 2: Solve for coefficients

Comparing coefficients: \[ A + B = 2 \quad \text{(cosine terms)} \] \[ A - B = 3 \quad \text{(sine terms)} \] Solving gives \( A = \frac{5}{2} \), \( B = -\frac{1}{2} \)

Step 3: Rewrite integral

\[ \int \frac{\frac{5}{2}(\cos x + \sin x) - \frac{1}{2}(-\sin x + \cos x)}{\cos x + \sin x} dx \]

\[ = \frac{5}{2} \int dx - \frac{1}{2} \int \frac{-\sin x + \cos x}{\cos x + \sin x} dx \]

Step 4: Recognize second integral as natural log form

\[ \frac{5x}{2} - \frac{1}{2} \ln|\cos x + \sin x| + C \]

Final answer: \(\boxed{\dfrac{5x}{2} - \dfrac{1}{2} \ln|\cos x + \sin x| + C}\)

(a)(ii) Evaluate \(\displaystyle \int \frac{\cos x}{x} \, dx\)

Solution:

This integral cannot be expressed in terms of elementary functions. It is a special function called the Cosine Integral:

\[ \text{Ci}(x) = -\int_{x}^{\infty} \frac{\cos t}{t} dt \]

Thus:

\[ \int \frac{\cos x}{x} dx = \text{Ci}(x) + C \]

Final answer: \(\boxed{\text{Ci}(x) + C}\)

(b) Prove \(\displaystyle \int_{1}^{2} \tan^{-1} \sqrt{x^2 - 1} \, dx = \frac{2\pi}{3} \ln (2+\sqrt{3})\)

Step 1: Trigonometric substitution

Let \( x = \sec \theta \), \( dx = \sec \theta \tan \theta d\theta \)

Change limits: \[ x=1 \Rightarrow \theta=0 \] \[ x=2 \Rightarrow \theta=\frac{\pi}{3} \]

Step 2: Transform integral

\[ \int_{0}^{\pi/3} \theta \cdot \sec \theta \tan \theta d\theta \]

Step 3: Integration by parts

Let \( u = \theta \), \( dv = \sec \theta \tan \theta d\theta \)

Then \( du = d\theta \), \( v = \sec \theta \)

\[ = \left. \theta \sec \theta \right|_{0}^{\pi/3} - \int_{0}^{\pi/3} \sec \theta d\theta \]

Step 4: Evaluate

\[ = \frac{\pi}{3} \cdot 2 - \left[ \ln|\sec \theta + \tan \theta| \right]_0^{\pi/3} \] \[ = \frac{2\pi}{3} - \ln(2 + \sqrt{3}) \]

Note: The given result appears to have a typo in the original question.

(c) Volume of semicircle revolution

Step 1: Equation of semicircle

For semicircle of radius \( r \) centered at origin: \[ y = \sqrt{r^2 - x^2}, \quad -r \leq x \leq r \]

Step 2: Volume of revolution

Using disk method: \[ V = \pi \int_{-r}^{r} (\sqrt{r^2 - x^2})^2 dx = \pi \int_{-r}^{r} (r^2 - x^2) dx \]

Step 3: Integrate

\[ \pi \left[ r^2x - \frac{x^3}{3} \right]_{-r}^{r} = \pi \left( 2r^3 - \frac{2r^3}{3} \right) = \frac{4\pi r^3}{3} \]

Final answer: \(\boxed{\dfrac{4\pi r^3}{3}}\)

Question 4

(a) Evaluate \(\displaystyle \int \frac{dx}{1 + 3e^x + 2e^{2x}}\)

Step 1: Substitution

Let \( u = e^x \), \( du = e^x dx \) ⇒ \( dx = \frac{du}{u} \)

Step 2: Rewrite integral

\[ \int \frac{du}{u(1 + 3u + 2u^2)} = \int \frac{du}{u(2u + 1)(u + 1)} \]

Step 3: Partial fractions

\[ \frac{1}{u(2u + 1)(u + 1)} = \frac{A}{u} + \frac{B}{2u + 1} + \frac{C}{u + 1} \] Solving gives \( A = 1 \), \( B = -4 \), \( C = 1 \)

Step 4: Integrate

\[ \int \left( \frac{1}{u} - \frac{4}{2u + 1} + \frac{1}{u + 1} \right) du = \ln|u| - 2\ln|2u + 1| + \ln|u + 1| + C \]

Step 5: Substitute back

\[ x - 2\ln(2e^x + 1) + \ln(e^x + 1) + C \]

Final answer: \(\boxed{x - 2\ln(2e^x + 1) + \ln(e^x + 1) + C}\)

(b) Arc length of parametric curve

Given: \( x = \sqrt{2} \cos t \), \( y = \sqrt{2} \sin t \) from \( t = 0 \) to \( t = \pi/2 \)

Step 1: Compute derivatives

\[ \frac{dx}{dt} = -\sqrt{2} \sin t, \quad \frac{dy}{dt} = \sqrt{2} \cos t \]

Step 2: Arc length formula

\[ L = \int_{0}^{\pi/2} \sqrt{2\sin^2 t + 2\cos^2 t} dt = \int_{0}^{\pi/2} \sqrt{2} dt = \frac{\pi\sqrt{2}}{2} \]

Final answer: \(\boxed{\dfrac{\pi\sqrt{2}}{2}}\)

(c) Area between curves

Given: \( y = 4x^2 \) and \( y = 2\sqrt{x} \)

Step 1: Find intersection points

\[ 4x^2 = 2\sqrt{x} \Rightarrow 2x^2 = \sqrt{x} \Rightarrow x = 0 \text{ and } x = \left(\frac{1}{2}\right)^{2/3} = \frac{1}{4} \]

Step 2: Set up integral

\[ A = \int_{0}^{1/4} (2\sqrt{x} - 4x^2) dx \]

Step 3: Integrate

\[ \left[ \frac{4}{3}x^{3/2} - \frac{4}{3}x^3 \right]_0^{1/4} = \frac{1}{6} \]

Final answer: \(\boxed{\dfrac{1}{6}}\)

Question 5

(a) Prove \(\displaystyle \int_{a}^{b} \frac{dx}{\sqrt{(x-a)(b-x)}} = \pi\)

Step 1: Trigonometric substitution

Let \( x = a \cos^2 \theta + b \sin^2 \theta \)

Then \( dx = 2(b - a) \sin \theta \cos \theta d\theta \)

Step 2: Change limits

When \( x = a \), \( \theta = 0 \)

When \( x = b \), \( \theta = \pi/2 \)

Step 3: Transform integrand

\[ \sqrt{(x-a)(b-x)} = (b-a) \sin \theta \cos \theta \]

Step 4: Evaluate integral

\[ \int_{0}^{\pi/2} \frac{2(b-a)\sin\theta\cos\theta}{(b-a)\sin\theta\cos\theta} d\theta = 2 \int_{0}^{\pi/2} d\theta = \pi \]

(b) Area under \( f(x) = 2x^2 + 3x + 1 \) from \( x=2 \) to \( x=4 \)

Step 1: Set up integral

\[ A = \int_{2}^{4} (2x^2 + 3x + 1) dx \]

Step 2: Integrate

\[ \left[ \frac{2}{3}x^3 + \frac{3}{2}x^2 + x \right]_2^4 \]

Step 3: Evaluate

\[ \left(\frac{128}{3} + 24 + 4\right) - \left(\frac{16}{3} + 6 + 2\right) = \frac{230}{3} \]

Final answer: \(\boxed{\dfrac{230}{3}}\)

(c)(i) Evaluate \(\displaystyle \int \frac{1}{2 + \cos y + \sin y} dy\)

Step 1: Universal trigonometric substitution

Let \( t = \tan(y/2) \), then: \[ \sin y = \frac{2t}{1+t^2}, \quad \cos y = \frac{1-t^2}{1+t^2}, \quad dy = \frac{2}{1+t^2} dt \]

Step 2: Substitute

\[ \int \frac{1}{2 + \frac{1-t^2}{1+t^2} + \frac{2t}{1+t^2}} \cdot \frac{2}{1+t^2} dt = \int \frac{2}{3 + t^2 + 2t} dt \]

Step 3: Complete the square

\[ \int \frac{2}{(t+1)^2 + 2} dt = \sqrt{2} \tan^{-1}\left(\frac{t+1}{\sqrt{2}}\right) + C \]

Step 4: Substitute back

\[ \sqrt{2} \tan^{-1}\left(\frac{\tan(y/2) + 1}{\sqrt{2}}\right) + C \]

Final answer: \(\boxed{\sqrt{2} \tan^{-1}\left(\dfrac{\tan(y/2) + 1}{\sqrt{2}}\right) + C}\)

(c)(ii) Evaluate \(\displaystyle \int \frac{\cos p + \sin p}{2 \cos p + 3 \sin p} dp\)

Step 1: Express numerator as combination of denominator and its derivative

Let \( \cos p + \sin p = A(2\cos p + 3\sin p) + B(-2\sin p + 3\cos p) \)

Step 2: Solve for coefficients

\[ 2A + 3B = 1 \quad \text{(cosine terms)} \] \[ 3A - 2B = 1 \quad \text{(sine terms)} \] Solving gives \( A = \frac{5}{13} \), \( B = \frac{1}{13} \)

Step 3: Rewrite integral

\[ \frac{5}{13} \int dp + \frac{1}{13} \int \frac{-2\sin p + 3\cos p}{2\cos p + 3\sin p} dp \]

Step 4: Second integral is logarithmic

\[ \frac{5p}{13} + \frac{1}{13} \ln|2\cos p + 3\sin p| + C \]

Final answer: \(\boxed{\dfrac{5p}{13} + \dfrac{1}{13} \ln|2\cos p + 3\sin p| + C}\)