Samacheer Kalvi Class 12 Business Maths Solutions Chapter 2 Integral Calculus I Exercise 2.9

Evaluate the following using properties of definite integrals:

Question 1. \( \int_{-\pi/4}^{\pi/4} x^3 \cos^3 x \, dx \)
Answer:
Let \( f(x) = x^3 \cos^3 x \)
Now, let's find \( f(-x) \):
\( f(-x) = (-x)^3 \cos^3 (-x) \)
Since \( (-x)^3 = -x^3 \) and \( \cos(-x) = \cos x \), we have:
\( f(-x) = -x^3 \cos^3 x \)
So, \( f(-x) = -f(x) \). This means \( f(x) \) is an odd function.
When an odd function is integrated over a symmetric interval (from \( -a \) to \( a \)), the result is always zero. This is a key property of definite integrals.
\( \therefore \int_{-\pi/4}^{\pi/4} x^3 \cos^3 x \, dx = 0 \)
In simple words: First, we check if the function is odd or even. If \( f(-x) \) is the same as \( -f(x) \), it's an odd function. For odd functions integrated from \( -a \) to \( a \), the answer is always zero, which makes solving easy.

🎯 Exam Tip: Always check if the integrand is an odd or even function first when the limits of integration are symmetric (from \( -a \) to \( a \)), as this can simplify the problem significantly to zero.

Question 2. \( \int_{-\pi/2}^{\pi/2} \sin^2 \theta \, d\theta \)
Answer:
Let \( f(\theta) = \sin^2 \theta \)
Now, let's find \( f(-\theta) \):
\( f(-\theta) = \sin^2 (-\theta) \)
We know that \( \sin(-\theta) = -\sin\theta \). So,
\( f(-\theta) = (-\sin\theta)^2 \)
\( f(-\theta) = \sin^2 \theta \)
Since \( f(-\theta) = f(\theta) \), this means \( f(\theta) \) is an even function.
For an even function integrated over a symmetric interval (from \( -a \) to \( a \)), we can write it as twice the integral from \( 0 \) to \( a \).
\( \therefore \int_{-\pi/2}^{\pi/2} \sin^2 \theta \, d\theta = 2 \int_{0}^{\pi/2} \sin^2 \theta \, d\theta \)
We use the trigonometric identity: \( \sin^2 \theta = \frac{1 - \cos 2\theta}{2} \). This helps to simplify the integration.
\( = 2 \int_{0}^{\pi/2} \frac{1 - \cos 2\theta}{2} \, d\theta \)
\( = \int_{0}^{\pi/2} (1 - \cos 2\theta) \, d\theta \)
Now, we integrate term by term:
\( = [\theta - \frac{\sin 2\theta}{2}]_{0}^{\pi/2} \)
Next, we apply the limits of integration:
\( = (\frac{\pi}{2} - \frac{\sin (2 \times \pi/2)}{2}) - (0 - \frac{\sin (2 \times 0)}{2}) \)
\( = (\frac{\pi}{2} - \frac{\sin \pi}{2}) - (0 - \frac{\sin 0}{2}) \)
Since \( \sin \pi = 0 \) and \( \sin 0 = 0 \):
\( = (\frac{\pi}{2} - 0) - (0 - 0) \)
\( = \frac{\pi}{2} \)
In simple words: We first check if the function is even. If \( f(-\theta) \) is the same as \( f(\theta) \), it's even. For even functions, we can integrate from \( 0 \) to \( a \) and double the answer. Then, we change \( \sin^2 \theta \) using a special math rule to make it easier to integrate and solve.

🎯 Exam Tip: When integrating even functions over symmetric intervals, simplify by changing the limits to \( 0 \) to \( a \) and multiplying by 2. Remember to use appropriate trigonometric identities to make the integral solvable, such as for \( \sin^2 \theta \) or \( \cos^2 \theta \).

Question 3. \( \int_{-1}^{1} \log(\frac { 2-x }{2+x}) \, dx \)
Answer:
Let \( f(x) = \log(\frac { 2-x }{2+x}) \)
Now, let's find \( f(-x) \):
\( f(-x) = \log(\frac { 2-(-x) }{2+(-x)}) \)
\( f(-x) = \log(\frac { 2+x }{2-x}) \)
Using logarithm properties, \( \log(\frac{a}{b}) = -\log(\frac{b}{a}) \). This property helps to switch the numerator and denominator.
\( f(-x) = -\log(\frac { 2-x }{2+x}) \)
So, \( f(-x) = -f(x) \). This means \( f(x) \) is an odd function.
For an odd function integrated over a symmetric interval (from \( -a \) to \( a \)), the integral is always zero.
\( \therefore \int_{-1}^{1} \log(\frac { 2-x }{2+x}) \, dx = 0 \)
In simple words: We check if the function is odd by putting \( -x \) instead of \( x \). If the new function is the negative of the original, it's an odd function. For odd functions, if you integrate them from a number to its negative (like -1 to 1), the answer is always zero.

🎯 Exam Tip: Always identify if a function is odd or even when integrating over a symmetric interval \( [-a, a] \). For odd functions, the integral is immediately 0, saving calculation time. Remember \( \log(\frac{A}{B}) = -\log(\frac{B}{A}) \) for checking odd/even nature.

Question 4. \( \int_{0}^{\pi/2} \frac { \sin^7x }{\sin^7x+\cos^7x} \, dx \)
Answer:
Let \( I = \int_{0}^{\pi/2} \frac { \sin^7x }{\sin^7x+\cos^7x} \, dx \) ...(1)
Using the property of definite integrals: \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a – x) \, dx \). This property is very useful for integrals with symmetric limits like \( 0 \) to \( a \).
So, we replace \( x \) with \( \frac{\pi}{2} - x \):
\( I = \int_{0}^{\pi/2} \frac { \sin^7(\pi/2-x) }{\sin^7(\pi/2-x)+\cos^7(\pi/2-x)} \, dx \)
We know that \( \sin(\pi/2-x) = \cos x \) and \( \cos(\pi/2-x) = \sin x \). Applying these, we get:
\( I = \int_{0}^{\pi/2} \frac { \cos^7x }{\cos^7x+\sin^7x} \, dx \) ...(2)
Now, we add equation (1) and equation (2):
\( I + I = \int_{0}^{\pi/2} \frac { \sin^7x }{\sin^7x+\cos^7x} \, dx + \int_{0}^{\pi/2} \frac { \cos^7x }{\cos^7x+\sin^7x} \, dx \)
Since the denominators are the same, we can combine the numerators:
\( 2I = \int_{0}^{\pi/2} \frac { \sin^7x + \cos^7x }{\sin^7x+\cos^7x} \, dx \)
\( 2I = \int_{0}^{\pi/2} 1 \, dx \)
Now, integrate 1 with respect to \( x \):
\( 2I = [x]_{0}^{\pi/2} \)
Apply the limits of integration:
\( 2I = \frac{\pi}{2} - 0 \)
\( 2I = \frac{\pi}{2} \)
Finally, solve for \( I \):
\( I = \frac{\pi}{4} \)
In simple words: We start with the integral and use a special rule that lets us replace \( x \) with \( a-x \) (here, \( \pi/2-x \)). After this, \( \sin \) becomes \( \cos \) and \( \cos \) becomes \( \sin \). Then, we add the original integral and the new one. The top and bottom parts become the same, so the fraction turns into 1, which is easy to integrate.

🎯 Exam Tip: This type of integral (where \( f(x) + f(a-x) \) simplifies the numerator) is a common application of the property \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a – x) \, dx \). Always look for opportunities to use this property when the limits are from \( 0 \) to \( a \).

Question 5. \( \int_{0}^{1} \log (\frac { 1 }{x} - 1) \, dx \)
Answer:
Let \( I = \int_{0}^{1} \log (\frac { 1 }{x} - 1) \, dx \)
First, simplify the term inside the logarithm:
\( I = \int_{0}^{1} \log (\frac { 1-x }{x}) \, dx \) ...(1)
Using the property of definite integrals: \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a – x) \, dx \). This property is key for problems with limits \( 0 \) to \( a \).
Here, \( a=1 \), so we replace \( x \) with \( 1-x \):
\( I = \int_{0}^{1} \log (\frac { 1-(1-x) }{1-x}) \, dx \)
Simplify the expression:
\( I = \int_{0}^{1} \log (\frac { x }{1-x}) \, dx \) ...(2)
Now, we add equation (1) and equation (2):
\( I + I = \int_{0}^{1} \log (\frac { 1-x }{x}) \, dx + \int_{0}^{1} \log (\frac { x }{1-x}) \, dx \)
Using the logarithm property \( \log A + \log B = \log (AB) \). This helps combine the two logarithms into one.
\( 2I = \int_{0}^{1} \log \left( \frac { 1-x }{x} \times \frac { x }{1-x} \right) \, dx \)
\( 2I = \int_{0}^{1} \log (1) \, dx \)
Since \( \log(1) = 0 \):
\( 2I = \int_{0}^{1} 0 \, dx \)
\( 2I = 0 \)
\( \therefore I = 0 \)
In simple words: We first rewrite the inside of the logarithm. Then, we use a special integral rule to replace \( x \) with \( 1-x \). When we add the original integral and the new one, the parts inside the logarithm multiply together to give 1. Since \( \log(1) \) is always zero, the whole integral becomes zero.

🎯 Exam Tip: When the integrand involves logarithms and the limits are from \( 0 \) to \( a \), consider using the property \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a – x) \, dx \). This often leads to terms that cancel out or simplify via \( \log A + \log B = \log(AB) \).

Question 6. \( \int_{0}^{1} \frac { x }{(1-x)^{3/4}} \, dx \)
Answer:
Let \( I = \int_{0}^{1} \frac { x }{(1-x)^{3/4}} \, dx \)
Using the property of definite integrals: \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a – x) \, dx \). This property is very useful for limits \( 0 \) to \( a \).
Here, \( a=1 \), so we replace \( x \) with \( 1-x \):
\( I = \int_{0}^{1} \frac { (1-x) }{(1-(1-x))^{3/4}} \, dx \)
Simplify the denominator:
\( I = \int_{0}^{1} \frac { (1-x) }{(x)^{3/4}} \, dx \)
Now, split the numerator and simplify the powers of \( x \):
\( I = \int_{0}^{1} (x^{-3/4} - x \cdot x^{-3/4}) \, dx \)
\( I = \int_{0}^{1} (x^{-3/4} - x^{1-3/4}) \, dx \)
\( I = \int_{0}^{1} (x^{-3/4} - x^{1/4}) \, dx \)
Next, integrate each term using the power rule \( \int x^n \, dx = \frac{x^{n+1}}{n+1} \). This is a standard integration technique.
\( I = \left[ \frac{x^{-3/4+1}}{-3/4+1} - \frac{x^{1/4+1}}{1/4+1} \right]_{0}^{1} \)
\( I = \left[ \frac{x^{1/4}}{1/4} - \frac{x^{5/4}}{5/4} \right]_{0}^{1} \)
\( I = \left[ 4x^{1/4} - \frac{4}{5}x^{5/4} \right]_{0}^{1} \)
Now, apply the limits of integration (upper limit minus lower limit):
\( I = (4(1)^{1/4} - \frac{4}{5}(1)^{5/4}) - (4(0)^{1/4} - \frac{4}{5}(0)^{5/4}) \)
\( I = (4 - \frac{4}{5}) - (0 - 0) \)
\( I = 4 - \frac{4}{5} \)
To subtract, find a common denominator:
\( I = \frac{20}{5} - \frac{4}{5} \)
\( I = \frac{16}{5} \)
In simple words: We use a rule to change \( x \) to \( 1-x \) in the integral. This helps simplify the bottom part of the fraction. After simplifying, we separate the terms and use the basic power rule to integrate each part. Finally, we put in the start and end numbers to get the answer.

🎯 Exam Tip: For integrals from \( 0 \) to \( 1 \), applying the property \( \int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a – x) \, dx \) often transforms the integrand into a simpler form that can be integrated using basic power rules.