Samacheer Kalvi Class 12 Maths Solutions Chapter 9 Applications of Integration Exercise 9.4

Question 1. Evaluate the following \( \int_{0}^{1} x^3 e^{-2x} \,dx \)
Answer: We will solve this integral using Bernoulli's formula, which helps integrate products of functions. This method simplifies the integration process for certain types of functions like polynomials multiplied by exponentials.
Given integral: \( I = \int_{0}^{1} x^3 e^{-2x} \,dx \)
Using Bernoulli's formula for integration by parts:
\( \int uv \,dx = uv_1 - u'v_2 + u''v_3 - u'''v_4 + \dots \)
Here, let \( u = x^3 \) and \( dv = e^{-2x} \,dx \).
We find the derivatives of \( u \) and integrals of \( v \):
\( u = x^3 \)
\( u' = 3x^2 \)
\( u'' = 6x \)
\( u''' = 6 \)
\( u'''' = 0 \)
And for \( v \):
\( v = e^{-2x} \)
\( v_1 = \int e^{-2x} \,dx = \frac{e^{-2x}}{-2} \)
\( v_2 = \int \frac{e^{-2x}}{-2} \,dx = \frac{e^{-2x}}{4} \)
\( v_3 = \int \frac{e^{-2x}}{4} \,dx = \frac{e^{-2x}}{-8} \)
\( v_4 = \int \frac{e^{-2x}}{-8} \,dx = \frac{e^{-2x}}{16} \)
Now, substitute these into Bernoulli's formula:
\( \int_{0}^{1} x^3 e^{-2x} \,dx = \left[ x^3 \left(\frac{e^{-2x}}{-2}\right) - (3x^2) \left(\frac{e^{-2x}}{4}\right) + (6x) \left(\frac{e^{-2x}}{-8}\right) - (6) \left(\frac{e^{-2x}}{16}\right) \right]_{0}^{1} \)
\( = \left[ -\frac{x^3 e^{-2x}}{2} - \frac{3x^2 e^{-2x}}{4} - \frac{6x e^{-2x}}{8} - \frac{6 e^{-2x}}{16} \right]_{0}^{1} \)
Simplify the coefficients:
\( = \left[ -\frac{x^3 e^{-2x}}{2} - \frac{3x^2 e^{-2x}}{4} - \frac{3x e^{-2x}}{4} - \frac{3 e^{-2x}}{8} \right]_{0}^{1} \)
Factor out \( e^{-2x} \):
\( = \left[ e^{-2x} \left( -\frac{x^3}{2} - \frac{3x^2}{4} - \frac{3x}{4} - \frac{3}{8} \right) \right]_{0}^{1} \)
Now, evaluate at the limits \( x=1 \) and \( x=0 \):
At \( x=1 \):
\( e^{-2(1)} \left( -\frac{1^3}{2} - \frac{3(1)^2}{4} - \frac{3(1)}{4} - \frac{3}{8} \right) = e^{-2} \left( -\frac{1}{2} - \frac{3}{4} - \frac{3}{4} - \frac{3}{8} \right) \)
\( = e^{-2} \left( -\frac{4}{8} - \frac{6}{8} - \frac{6}{8} - \frac{3}{8} \right) = e^{-2} \left( -\frac{4+6+6+3}{8} \right) = e^{-2} \left( -\frac{19}{8} \right) \)
At \( x=0 \):
\( e^{-2(0)} \left( -\frac{0^3}{2} - \frac{3(0)^2}{4} - \frac{3(0)}{4} - \frac{3}{8} \right) = e^0 \left( 0 - 0 - 0 - \frac{3}{8} \right) = 1 \left( -\frac{3}{8} \right) = -\frac{3}{8} \)
Subtract the lower limit from the upper limit:
\( I = \left( e^{-2} \left( -\frac{19}{8} \right) \right) - \left( -\frac{3}{8} \right) \)
\( I = -\frac{19e^{-2}}{8} + \frac{3}{8} \)
\( I = \frac{3 - 19e^{-2}}{8} \)
The final answer is \( \frac{3 - 19e^{-2}}{8} \).
In simple words: We used a special integration method called Bernoulli's formula to solve this problem. We found the derivatives of the polynomial part and the integrals of the exponential part, then plugged them into the formula. Finally, we put in the upper and lower limits to get the numerical answer.

🎯 Exam Tip: Remember to carefully apply the signs in Bernoulli's formula \( (+, -, +, -) \) and perform the evaluation at the limits of integration very accurately to avoid errors.

Question 2. Evaluate \( \int_{0}^{1} \frac{\sin(3\tan^{-1}x)\tan^{-1}x}{1+x^2} \,dx \)
Answer: We need to evaluate the given definite integral. We will start by using a substitution to simplify the integral.
Let \( I = \int_{0}^{1} \frac{\sin(3\tan^{-1}x)\tan^{-1}x}{1+x^2} \,dx \)
Let \( t = \tan^{-1}x \).
Then, differentiate \( t \) with respect to \( x \): \( dt = \frac{1}{1+x^2} \,dx \).
Next, change the limits of integration based on the substitution:
When \( x = 0 \), \( t = \tan^{-1}(0) = 0 \).
When \( x = 1 \), \( t = \tan^{-1}(1) = \frac{\pi}{4} \).
Substitute \( t \) and \( dt \) into the integral:
\( I = \int_{0}^{\pi/4} (\sin(3t))t \,dt \)
Rearrange the terms:
\( I = \int_{0}^{\pi/4} t \sin(3t) \,dt \)
Now, we use integration by parts for \( \int t \sin(3t) \,dt \). We'll use the tabular method (or Bernoulli's formula for two terms):
Let \( u = t \) and \( dv = \sin(3t) \,dt \).
Then \( u' = 1 \) and \( u'' = 0 \).
And \( v_1 = \int \sin(3t) \,dt = -\frac{\cos(3t)}{3} \).
\( v_2 = \int -\frac{\cos(3t)}{3} \,dt = -\frac{\sin(3t)}{9} \).
Applying the integration by parts formula \( \int uv \,dx = uv_1 - u'v_2 \):
\( I = \left[ t \left(-\frac{\cos(3t)}{3}\right) - (1) \left(-\frac{\sin(3t)}{9}\right) \right]_{0}^{\pi/4} \)
\( I = \left[ -\frac{t\cos(3t)}{3} + \frac{\sin(3t)}{9} \right]_{0}^{\pi/4} \)
Now, evaluate at the limits \( t=\frac{\pi}{4} \) and \( t=0 \):
At \( t=\frac{\pi}{4} \):
\( -\frac{(\pi/4)\cos(3\pi/4)}{3} + \frac{\sin(3\pi/4)}{9} \)
Recall that \( \cos(3\pi/4) = -\frac{1}{\sqrt{2}} \) and \( \sin(3\pi/4) = \frac{1}{\sqrt{2}} \).
\( = -\frac{(\pi/4)(-1/\sqrt{2})}{3} + \frac{1/\sqrt{2}}{9} \)
\( = \frac{\pi}{12\sqrt{2}} + \frac{1}{9\sqrt{2}} \)
At \( t=0 \):
\( -\frac{(0)\cos(0)}{3} + \frac{\sin(0)}{9} = 0 + 0 = 0 \)
Subtract the lower limit from the upper limit:
\( I = \left( \frac{\pi}{12\sqrt{2}} + \frac{1}{9\sqrt{2}} \right) - 0 \)
Combine the terms by finding a common denominator for the coefficients (LCM of 12 and 9 is 36):
\( I = \frac{3\pi}{36\sqrt{2}} + \frac{4}{36\sqrt{2}} \)
\( I = \frac{3\pi + 4}{36\sqrt{2}} \)
This is equivalent to the form \( \frac{1}{\sqrt{2}} \left( \frac{\pi}{12} + \frac{1}{9} \right) \). This result involves careful application of trigonometric values.
In simple words: We first changed the variable in the integral to make it simpler. Then we used a method called integration by parts, which is like a reverse product rule for differentiation. After that, we put in the starting and ending values to get our final numerical answer.

🎯 Exam Tip: When using substitution, always remember to change the limits of integration according to the new variable. Forgetting this is a common mistake.

Question 3. Evaluate \( \int_{0}^{1/\sqrt{2}} \frac{e^{\sin^{-1}x}\sin^{-1}x}{\sqrt{1-x^2}} \,dx \)
Answer: We will evaluate this definite integral by using a suitable substitution.
Let \( I = \int_{0}^{1/\sqrt{2}} \frac{e^{\sin^{-1}x}\sin^{-1}x}{\sqrt{1-x^2}} \,dx \)
Let \( t = \sin^{-1}x \).
Differentiate \( t \) with respect to \( x \): \( dt = \frac{1}{\sqrt{1-x^2}} \,dx \).
Now, change the limits of integration for \( t \):
When \( x = 0 \), \( t = \sin^{-1}(0) = 0 \).
When \( x = \frac{1}{\sqrt{2}} \), \( t = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4} \).
Substitute \( t \) and \( dt \) into the integral:
\( I = \int_{0}^{\pi/4} e^t t \,dt \)
Rearrange the terms:
\( I = \int_{0}^{\pi/4} t e^t \,dt \)
Now, we use integration by parts for \( \int t e^t \,dt \).
Let \( u = t \) and \( dv = e^t \,dt \).
Then \( u' = 1 \) and \( u'' = 0 \).
And \( v_1 = \int e^t \,dt = e^t \).
\( v_2 = \int e^t \,dt = e^t \).
Applying the integration by parts formula \( \int uv \,dx = uv_1 - u'v_2 \):
\( I = [te^t - (1)e^t]_{0}^{\pi/4} \)
\( I = [e^t(t-1)]_{0}^{\pi/4} \)
Now, evaluate at the limits \( t=\frac{\pi}{4} \) and \( t=0 \):
At \( t=\frac{\pi}{4} \):
\( e^{\pi/4} \left(\frac{\pi}{4} - 1\right) \)
At \( t=0 \):
\( e^0 (0-1) = 1(-1) = -1 \)
Subtract the lower limit from the upper limit:
\( I = e^{\pi/4} \left(\frac{\pi}{4} - 1\right) - (-1) \)
\( I = e^{\pi/4} \left(\frac{\pi}{4} - 1\right) + 1 \)
This expression gives the exact value of the integral. The combination of substitution and integration by parts is very effective for these types of problems.
In simple words: First, we swapped a complex part of the integral for a simpler variable. This also changed the starting and ending points of our integration. Then, we used a special rule called integration by parts to solve the new integral. Finally, we put back the numbers for the start and end points to get our answer.

🎯 Exam Tip: Always look for suitable substitutions that simplify the integrand and the differential term. This is often the first and most critical step in solving complex integrals.

Question 4. Evaluate \( \int_{0}^{\pi/2} x^2 \cos(2x) \,dx \)
Answer: We need to evaluate this definite integral using integration by parts. Since \( x^2 \) is a polynomial and \( \cos(2x) \) is a trigonometric function, Bernoulli's formula is a good choice.
Let \( I = \int_{0}^{\pi/2} x^2 \cos(2x) \,dx \)
Using Bernoulli's formula \( \int uv \,dx = uv_1 - u'v_2 + u''v_3 - \dots \):
Let \( u = x^2 \) and \( dv = \cos(2x) \,dx \).
Derivatives of \( u \):
\( u = x^2 \)
\( u' = 2x \)
\( u'' = 2 \)
\( u''' = 0 \)
Integrals of \( v \):
\( v_1 = \int \cos(2x) \,dx = \frac{\sin(2x)}{2} \)
\( v_2 = \int \frac{\sin(2x)}{2} \,dx = \frac{1}{2} \left(-\frac{\cos(2x)}{2}\right) = -\frac{\cos(2x)}{4} \)
\( v_3 = \int -\frac{\cos(2x)}{4} \,dx = -\frac{1}{4} \left(\frac{\sin(2x)}{2}\right) = -\frac{\sin(2x)}{8} \)
Apply Bernoulli's formula:
\( I = \left[ x^2 \left(\frac{\sin(2x)}{2}\right) - (2x) \left(-\frac{\cos(2x)}{4}\right) + (2) \left(-\frac{\sin(2x)}{8}\right) \right]_{0}^{\pi/2} \)
Simplify the expression:
\( I = \left[ \frac{x^2 \sin(2x)}{2} + \frac{2x \cos(2x)}{4} - \frac{2 \sin(2x)}{8} \right]_{0}^{\pi/2} \)
\( I = \left[ \frac{x^2 \sin(2x)}{2} + \frac{x \cos(2x)}{2} - \frac{\sin(2x)}{4} \right]_{0}^{\pi/2} \)
Now, evaluate at the limits \( x=\frac{\pi}{2} \) and \( x=0 \):
At \( x=\frac{\pi}{2} \):
\( \sin\left(2 \cdot \frac{\pi}{2}\right) = \sin(\pi) = 0 \)
\( \cos\left(2 \cdot \frac{\pi}{2}\right) = \cos(\pi) = -1 \)
Substitute these values:
\( \frac{(\pi/2)^2 (0)}{2} + \frac{(\pi/2) (-1)}{2} - \frac{(0)}{4} \)
\( = 0 - \frac{\pi}{4} - 0 = -\frac{\pi}{4} \)
At \( x=0 \):
\( \sin(2 \cdot 0) = \sin(0) = 0 \)
\( \cos(2 \cdot 0) = \cos(0) = 1 \)
Substitute these values:
\( \frac{(0)^2 (0)}{2} + \frac{(0) (1)}{2} - \frac{(0)}{4} \)
\( = 0 + 0 - 0 = 0 \)
Subtract the lower limit from the upper limit:
\( I = \left( -\frac{\pi}{4} \right) - (0) \)
\( I = -\frac{\pi}{4} \)
This demonstrates how powerful integration by parts is for solving integrals of products of functions, especially when one function simplifies upon repeated differentiation.
In simple words: We used a special rule to integrate a multiplication of two functions, \( x^2 \) and \( \cos(2x) \). We found the derivatives of \( x^2 \) and the integrals of \( \cos(2x) \), then put them into the formula. Finally, we calculated the value at the start and end points to get the answer.

🎯 Exam Tip: Be careful with trigonometric values at special angles like \( \pi/2 \) and \( \pi \), and always double-check the signs when applying Bernoulli's formula or repeated integration by parts.