Use integration by parts to find $\, \int xe^{x}dx$ - HSC - SSCE Mathematics Extension 2 - Question 1 - 2004 - Paper 1

To evaluate $\int_{0}^{\frac{\pi}{2}} \frac{\sin x}{\cos^{3} x} \, dx$, we can use the substitution $u = \cos x$, thus:

• $du = -\sin x \, dx$
• The limits change: When $x = 0, u = 1$ and when $x = \frac{\pi}{2}, u = 0$.

This gives:

$$\int_{1}^{0} \frac{-1}{u^{3}} du = \int_{0}^{1} \frac{1}{u^{3}} du.$$

Calculating the integral:

$$= \left[-\frac{1}{2u^{2}}\right]_{0}^{1} = \left( -\frac{1}{2(1)^{2}} \right) - (0) = -\frac{1}{2}.$$

To complete the square for $5 + 4x - x^{2}$, we rewrite it as:

$$5 + 4x - x^{2} = -(x^{2} - 4x - 5) = -\left((x-2)^{2} - 9\right) = 9 - (x-2)^{2}.$$

Thus, we have:

$$\int \frac{dx}{\sqrt{9 - (x-2)^{2}}}.$$

Using the substitution $u = x - 2$ gives:

$$\int \frac{du}{\sqrt{9 - u^{2}}},$$

which results in:

$$= \sin^{-1} \left( \frac{u}{3} \right) + C = \sin^{-1} \left( \frac{x-2}{3} \right) + C.$$

To find $a$ and $b$, first equate the coefficients of the left and right hand sides after bringing everything over to one side, giving a common denominator of $(x+1)(x-1)^{2}$. Hence,

$\frac{x^{2}-7x+4}{(x+1)(x-1)^{2}} = \frac{a(x-1)^{2} + b(x+1)(x-1) - 1(x+1)}{(x+1)(x-1)^{2}}.$

Expanding and solving for $a$ and $b$, set coefficients equal:

1. $a + b - 1= 1$ (coefficient of $x^{2}$)
2. $-2a + b - 7 = -7$ (coefficient of $x^{1}$)
3. $a - b + 4 = 4$ (constant term)

Solving these simultaneously gives $a = 2$ and $b = 1$.

Using the values found previously ($a = 2$, $b = 1$), we rewrite the integral as:

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

Calculating each term individually, we have:

1. $\int \frac{2}{x+1} \, dx = 2 \ln |x+1| + C_{1}$
2. $\int \frac{1}{x-1} \, dx = \ln |x-1| + C_{2}$
3. $\int -\frac{1}{(x-1)^{2}} \, dx = \frac{1}{x-1} + C_{3}$

Combining the results:

$$\int \frac{x^{2}-7x+4}{(x+1)(x-1)^{2}} , dx = 2\ln |x+1| + \ln |x-1| + \frac{1}{x-1} + C.$

To solve the integral using the substitution $x = 2\sin\theta$, we have:

• $dx = 2\cos\theta \, d\theta$.
• Changing the limits accordingly:
  • When $x=0$, $\theta=0$; when $x=2$, $\theta=\frac{\pi}{2}$.

Substituting into the integral:

$$$$

Calculating gives:

$$= 2\left[\sin\theta\right]_{0}^{\frac{\pi}{2}} = 2(1 - 0) = 2.$