Samacheer Kalvi Class 12 Maths Solutions Chapter 9 Applications of Integration Exercise 9.5

Tamilnadu Samacheer Kalvi 12th Maths Solutions Chapter 9 Applications of Integration Ex 9.5

Question 1. Evaluate the following:

(i) \( \int_{0}^{\pi/2} \frac{dx}{1+5\cos^2x} \)
Answer:

To solve this integral, we first simplify the denominator. We replace 1 with \( \sin^2x + \cos^2x \).

\[ \int_{0}^{\pi/2} \frac{dx}{1+5\cos^2x} = \int_{0}^{\pi/2} \frac{dx}{\sin^2x + \cos^2x + 5\cos^2x} \]

\[ = \int_{0}^{\pi/2} \frac{dx}{\sin^2x + 6\cos^2x} \]

Next, we divide both the numerator and denominator by \( \cos^2x \) to make it easier to integrate.

\[ = \int_{0}^{\pi/2} \frac{\frac{dx}{\cos^2x}}{\frac{\sin^2x}{\cos^2x} + \frac{6\cos^2x}{\cos^2x}} = \int_{0}^{\pi/2} \frac{\sec^2x \ dx}{\tan^2x + 6} \]

Now, we use a substitution to simplify the integral further. Let \( u = \tan x \). Then, the derivative \( du = \sec^2x \ dx \).

When we change the variable from \( x \) to \( u \), the limits of integration also change:

\[ I = \int_{0}^{\infty} \frac{du}{u^2 + 6} \]

We can now integrate this using the standard formula for integrals of the form \( \int \frac{1}{x^2+a^2} dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \). Here, \( a = \sqrt{6} \).

\[ = \left[ \frac{1}{\sqrt{6}} \tan^{-1} \left( \frac{u}{\sqrt{6}} \right) \right]_{0}^{\infty} \]

Finally, we apply the upper and lower limits of integration. This type of integral is very common in physics, especially when dealing with oscillatory motion or electromagnetic fields.

\[ = \frac{1}{\sqrt{6}} (\tan^{-1}(\infty) - \tan^{-1}(0)) \]

\[ = \frac{1}{\sqrt{6}} \left( \frac{\pi}{2} - 0 \right) \]

\[ = \frac{\pi}{2\sqrt{6}} \]

In simple words: First, we made the bottom part of the fraction simpler. Then, we changed the variable and its limits to solve the integral. The final answer is \( \frac{\pi}{2\sqrt{6}} \).

🎯 Exam Tip: Remember to change the limits of integration carefully when you use a substitution method for definite integrals. A common mistake is forgetting this step.

(ii) \( \int_{0}^{\pi/2} \frac{dx}{5+4\sin^2x} \)
Answer:

To evaluate this integral, we first rewrite the denominator using the identity \( 1 = \sin^2x + \cos^2x \).

\[ I = \int_{0}^{\pi/2} \frac{dx}{5+4\sin^2x} = \int_{0}^{\pi/2} \frac{dx}{5(\sin^2x + \cos^2x) + 4\sin^2x} \]

\[ = \int_{0}^{\pi/2} \frac{dx}{5\cos^2x + 9\sin^2x} \]

Next, we divide the numerator and denominator by \( \cos^2x \) to prepare for substitution.

\[ = \int_{0}^{\pi/2} \frac{\frac{dx}{\cos^2x}}{\frac{5\cos^2x}{\cos^2x} + \frac{9\sin^2x}{\cos^2x}} = \int_{0}^{\pi/2} \frac{\sec^2x \ dx}{5 + 9\tan^2x} \]

We use a substitution where \( u = \tan x \), so \( du = \sec^2x \ dx \).

As before, the limits of integration change when we switch variables:

\[ I = \int_{0}^{\infty} \frac{du}{5 + 9u^2} \]

To use the standard integration formula, we factor out 9 from the denominator.

\[ = \frac{1}{9} \int_{0}^{\infty} \frac{du}{\frac{5}{9} + u^2} \]

Now, we apply the integral formula \( \int \frac{1}{a^2+x^2} dx = \frac{1}{a} \tan^{-1} \left( \frac{x}{a} \right) \) with \( a^2 = \frac{5}{9} \), so \( a = \frac{\sqrt{5}}{3} \). These types of integrals are often encountered when dealing with geometric problems involving areas or volumes under curves.

\[ = \frac{1}{9} \left[ \frac{1}{\frac{\sqrt{5}}{3}} \tan^{-1} \left( \frac{u}{\frac{\sqrt{5}}{3}} \right) \right]_{0}^{\infty} \]

\[ = \frac{1}{9} \left[ \frac{3}{\sqrt{5}} \tan^{-1} \left( \frac{3u}{\sqrt{5}} \right) \right]_{0}^{\infty} \]

Finally, we substitute the upper and lower limits into the expression.

\[ = \frac{1}{9} \cdot \frac{3}{\sqrt{5}} (\tan^{-1}(\infty) - \tan^{-1}(0)) \]

\[ = \frac{1}{3\sqrt{5}} \left( \frac{\pi}{2} - 0 \right) \]

\[ = \frac{\pi}{6\sqrt{5}} \]

In simple words: We first adjusted the bottom part of the fraction and then used a substitution. After that, we integrated the simplified expression and applied the new limits to get the final answer.

🎯 Exam Tip: When factoring constants out of the denominator, be careful to apply the factor to the entire denominator and outside the integral correctly.