SSCE HSC Mathematics Extension 2 Integration by substitution: Find $$\int

Find $$\int_0^5 \tan^3{x} \sec^2{x} \,dx.$$ (b) By completing the square, find $$\int \frac{dx}{\sqrt{-4+x^2+1}}.$$ (c) Use integration by parts to evaluate $$\int^4 \frac{\ln{x}}{x^2} \,dx.$$ (d) Use the substitution $u=\sqrt{x-1}$ to evaluate $$\int_2^3 \frac{1+x}{\sqrt{1-x}} \,dx.$$ (e) (i) Find real numbers $\alpha$ and $b$ such that $$\frac{5x^2-3x+1}{(x+1)(x-2)}=\frac{\alpha}{x^2+1}+\frac{b}{x-2}.$$ (ii) Find $$\int \frac{5x^2-3x+1}{(x^2+1)(x-2)} \,dx.$$ - HSC - SSCE Mathematics Extension 2 - Question 1 - 2001 - Paper 1

Question 1

$Find-$$\int_0^5-\tan^3{x}-\sec^2{x}-\,dx.$$--(b)-By-completing-the-square,-find-$$\int-\frac{dx}{\sqrt{-4+x^2+1}}.$$--(c)-Use-integration-by-parts-to-evaluate-$$\int^4-\frac{\ln{x}}{x^2}-\,dx.$$--(d)-Use-the-substitution-$u=\sqrt{x-1}$-to-evaluate-$$\int_2^3-\frac{1+x}{\sqrt{1-x}}-\,dx.$$--(e)-(i)-Find-real-numbers-$\alpha$-and-$b$-such-that-$$\frac{5x^2-3x+1}{(x+1)(x-2)}=\frac{\alpha}{x^2+1}+\frac{b}{x-2}.$$--(ii)-Find-$$\int-\frac{5x^2-3x+1}{(x^2+1)(x-2)}-\,dx.$$-HSC-SSCE Mathematics Extension 2-Question 1-2001-Paper 1.png$

Find $$\int_0^5 \tan^3{x} \sec^2{x} \,dx.$$ (b) By completing the square, find $$\int \frac{dx}{\sqrt{-4+x^2+1}}.$$ (c) Use integration by parts to evaluate $$\int... show full transcript

Worked Solution & Example Answer:Find $$\int_0^5 \tan^3{x} \sec^2{x} \,dx.$$ (b) By completing the square, find $$\int \frac{dx}{\sqrt{-4+x^2+1}}.$$ (c) Use integration by parts to evaluate $$\int^4 \frac{\ln{x}}{x^2} \,dx.$$ (d) Use the substitution $u=\sqrt{x-1}$ to evaluate $$\int_2^3 \frac{1+x}{\sqrt{1-x}} \,dx.$$ (e) (i) Find real numbers $\alpha$ and $b$ such that $$\frac{5x^2-3x+1}{(x+1)(x-2)}=\frac{\alpha}{x^2+1}+\frac{b}{x-2}.$$ (ii) Find $$\int \frac{5x^2-3x+1}{(x^2+1)(x-2)} \,dx.$$ - HSC - SSCE Mathematics Extension 2 - Question 1 - 2001 - Paper 1

Step 1

Find $ \int_0^5 \tan^3{x} \sec^2{x} \,dx $

96%

114 rated

Answer

This integral can be solved using the identity that relates tangent and secant. Start by noticing that $\sec^2{x} = 1 + \tan^2{x}.$ Thus, we can rewrite the integral as: $$\int_0^5 \tan^3{x} (1 + \tan^2{x}) ,dx.$ Split the integral into two parts and integrate accordingly.

Step 2

By completing the square, find $ \int \frac{dx}{\sqrt{-4+x^2+1}} $

99%

104 rated

Answer

To complete the square, rewrite the expression under the square root: $\sqrt{x^2 - 3}.$ This leads to the integral becoming: $\int \frac{dx}{\sqrt{(x-\sqrt{3})(x+\sqrt{3})}}.$ Apply a suitable trigonometric substitution to solve.

Step 3

Use integration by parts to evaluate $ \int^4 \frac{\ln{x}}{x^2} \,dx $

96%

101 rated

Answer

Let:

( u = \ln{x} )
( dv = \frac{1}{x^2} dx ) Then:
( du = \frac{1}{x} dx )
( v = -\frac{1}{x} ).

Apply integration by parts: $\int u \, dv = uv - \int v \, du.$

Step 4

Use the substitution $u=\sqrt{x-1}$ to evaluate $ \int_2^3 \frac{1+x}{\sqrt{1-x}} \,dx $

98%

120 rated

Answer

With the substitution, express in terms of ( u ):

Replace ( x = u^2 + 1 ) leading to changes in the limits of integration. The integral now simplifies, allowing us to resolve it easily.

Step 5

Find real numbers $ \alpha $ and $ b $ such that $ \frac{5x^2-3x+1}{(x+1)(x-2)}=\frac{\alpha}{x^2+1}+\frac{b}{x-2} $

97%

117 rated

Answer

Perform partial fraction decomposition on the left-hand side, then equate coefficients of like terms to solve for ( \alpha ) and ( b ).

Step 6

Find $ \int \frac{5x^2-3x+1}{(x^2+1)(x-2)} \,dx $

97%

121 rated

Answer

Use the results from the partial fractions obtained previously. Break down the integral into simpler parts and integrate each term individually.

Find \( \int_0^5 \tan^3{x} \sec^2{x} \,dx \)

By completing the square, find \( \int \frac{dx}{\sqrt{-4+x^2+1}} \)

Use integration by parts to evaluate \( \int^4 \frac{\ln{x}}{x^2} \,dx \)

Use the substitution $u=\sqrt{x-1}$ to evaluate \( \int_2^3 \frac{1+x}{\sqrt{1-x}} \,dx \)

Find real numbers \( \alpha \) and \( b \) such that \( \frac{5x^2-3x+1}{(x+1)(x-2)}=\frac{\alpha}{x^2+1}+\frac{b}{x-2} \)

Find \( \int \frac{5x^2-3x+1}{(x^2+1)(x-2)} \,dx \)

Join the SSCE students using SimpleStudy...