Samacheer Kalvi Class 11 Maths Solutions Chapter 11 Integral Calculus Exercise 11.12

Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 11 Integral Calculus Ex 11.12

Integrate the Following with Respect to x.

Question 1. (i) \(\sqrt{x^{2}+2x+10}\)
Answer:
We need to integrate the given expression. First, we complete the square inside the square root to make it easier to work with.
\[ \int \sqrt{x^2 + 2x + 10} \, dx \]
\[ = \int \sqrt{(x+1)^2 - 1^2 + 10} \, dx \]
\[ = \int \sqrt{(x+1)^2 + 9} \, dx \]
\[ = \int \sqrt{(x+1)^2 + 3^2} \, dx \]
Now, let \( t = x+1 \). This means that \( dt = dx \).
\[ = \int \sqrt{t^2 + 3^2} \, dt \]
We use the standard integration formula for \( \int \sqrt{x^2 + a^2} \, dx = \frac{x}{2}\sqrt{x^2 + a^2} + \frac{a^2}{2} \log|x + \sqrt{x^2 + a^2}| + c \). This formula helps us integrate expressions that look like the square root of a sum of squares.
Applying this formula with \( t \) and \( a=3 \):
\[ = \frac{t}{2}\sqrt{t^2+3^2} + \frac{3^2}{2} \log|t + \sqrt{t^2+3^2}| + c \]
Finally, we replace \( t \) back with \( x+1 \):
\[ = \frac{x+1}{2}\sqrt{(x+1)^2+3^2} + \frac{9}{2} \log|x+1 + \sqrt{(x+1)^2+9}| + c \]
\[ = \frac{x+1}{2}\sqrt{x^2+2x+1+9} + \frac{9}{2} \log|x+1 + \sqrt{x^2+2x+1+9}| + c \]
\[ = \frac{x+1}{2}\sqrt{x^2+2x+10} + \frac{9}{2} \log|x+1 + \sqrt{x^2+2x+10}| + c \]
In simple words: We changed the expression inside the square root to a simpler form, then used a substitution to match it with a known integration formula, and finally substituted back to get the answer.

🎯 Exam Tip: Always remember to complete the square if the quadratic expression under the square root is not already in the form \( (x \pm a)^2 \pm b^2 \).

Question 1. (ii) \(\sqrt{x^{2}-2 x-3}\)
Answer:
To integrate this expression, we first complete the square inside the root sign.
\[ \int \sqrt{x^2 - 2x - 3} \, dx \]
\[ = \int \sqrt{(x-1)^2 - 1^2 - 3} \, dx \]
\[ = \int \sqrt{(x-1)^2 - 4} \, dx \]
\[ = \int \sqrt{(x-1)^2 - 2^2} \, dx \]
Let \( t = x-1 \). This means \( dt = dx \).
\[ = \int \sqrt{t^2 - 2^2} \, dt \]
We will use the standard integration formula for \( \int \sqrt{x^2 - a^2} \, dx = \frac{x}{2}\sqrt{x^2 - a^2} - \frac{a^2}{2} \log|x + \sqrt{x^2 - a^2}| + c \). This formula is used for integrals involving the square root of a difference of squares.
Applying this formula with \( t \) and \( a=2 \):
\[ = \frac{t}{2}\sqrt{t^2-2^2} - \frac{2^2}{2} \log|t + \sqrt{t^2-2^2}| + c \]
Now, substitute \( t \) back with \( x-1 \):
\[ = \frac{x-1}{2}\sqrt{(x-1)^2-2^2} - \frac{4}{2} \log|x-1 + \sqrt{(x-1)^2-4}| + c \]
\[ = \frac{x-1}{2}\sqrt{x^2-2x+1-4} - 2 \log|x-1 + \sqrt{x^2-2x+1-4}| + c \]
\[ = \frac{x-1}{2}\sqrt{x^2-2x-3} - 2 \log|x-1 + \sqrt{x^2-2x-3}| + c \]
In simple words: We rearranged the terms inside the square root, then used a substitution and a known formula to find the integral, finally putting the 'x' terms back.

🎯 Exam Tip: When dealing with \( \sqrt{x^2 - a^2} \), pay close attention to the minus sign in the logarithm term of the integration formula.

Question 1. (iii) \(\sqrt{(6-x)(x-4)}\)
Answer:
First, we need to expand the product inside the square root and then complete the square.
\[ \int \sqrt{(6-x)(x-4)} \, dx \]
\[ = \int \sqrt{6x-24-x^2+4x} \, dx \]
\[ = \int \sqrt{10x-x^2-24} \, dx \]
\[ = \int \sqrt{-24-(x^2-10x)} \, dx \]
To complete the square for \( x^2-10x \), we add and subtract \( (10/2)^2 = 25 \):
\[ = \int \sqrt{-24-[(x-5)^2-5^2]} \, dx \]
\[ = \int \sqrt{-24-(x-5)^2+25} \, dx \]
\[ = \int \sqrt{1-(x-5)^2} \, dx \]
Next, let \( t = x-5 \). This implies that \( dt = dx \).
\[ = \int \sqrt{1^2-t^2} \, dt \]
We will use the standard integration formula for \( \int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a}) + c \). This formula helps integrate expressions that are the square root of a constant squared minus a variable squared.
Applying this formula with \( t \) and \( a=1 \):
\[ = \frac{t}{2}\sqrt{1^2-t^2} + \frac{1^2}{2} \sin^{-1}(\frac{t}{1}) + c \]
Finally, substitute \( t \) back with \( x-5 \):
\[ = \frac{x-5}{2}\sqrt{1-(x-5)^2} + \frac{1}{2} \sin^{-1}(x-5) + c \]
\[ = \frac{x-5}{2}\sqrt{1-(x^2-10x+25)} + \frac{1}{2} \sin^{-1}(x-5) + c \]
\[ = \frac{x-5}{2}\sqrt{1-x^2+10x-25} + \frac{1}{2} \sin^{-1}(x-5) + c \]
\[ = \frac{x-5}{2}\sqrt{10x-x^2-24} + \frac{1}{2} \sin^{-1}(x-5) + c \]
In simple words: We first multiplied out the terms and then rearranged them to fit a known integration pattern. After using a substitution and a special formula, we put the original 'x' terms back to get the final answer.

🎯 Exam Tip: When the expression under the square root involves \( -x^2 \), factoring out the minus sign and completing the square is a common technique to transform it into the \( \sqrt{a^2-x^2} \) form.

Question 2. (i) \(\sqrt{9-(2 x+5)^{2}}\)
Answer:
We need to integrate this expression. Let's make a substitution to simplify it.
\[ \int \sqrt{9 - (2x+5)^2} \, dx \]
\[ = \int \sqrt{3^2 - (2x+5)^2} \, dx \]
Let \( t = 2x+5 \). When we differentiate both sides, we get \( 2 \, dx = dt \), which means that \( dx = \frac{1}{2} \, dt \).
\[ = \int \sqrt{3^2 - t^2} \frac{1}{2} \, dt \]
\[ = \frac{1}{2} \int \sqrt{3^2 - t^2} \, dt \]
We use the standard integration formula for \( \int \sqrt{a^2-x^2} \, dx = \frac{x}{2}\sqrt{a^2-x^2} + \frac{a^2}{2} \sin^{-1}(\frac{x}{a}) + c \). This formula is very helpful for integrals involving square roots of differences of squares.
Applying this formula with \( t \) and \( a=3 \):
\[ = \frac{1}{2} \left[ \frac{t}{2}\sqrt{3^2-t^2} + \frac{3^2}{2} \sin^{-1}\left(\frac{t}{3}\right) \right] + c \]
Finally, we substitute \( t \) back with \( 2x+5 \):
\[ = \frac{1}{2} \left[ \frac{2x+5}{2}\sqrt{3^2-(2x+5)^2} + \frac{9}{2} \sin^{-1}\left(\frac{2x+5}{3}\right) \right] + c \]
\[ = \frac{1}{4} (2x+5)\sqrt{9-(2x+5)^2} + \frac{9}{4} \sin^{-1}\left(\frac{2x+5}{3}\right) + c \]
In simple words: We simplified the integral by using a substitution, then applied a standard formula, and put back the original expression to get the final answer.

🎯 Exam Tip: Remember to adjust \( dx \) properly when making a substitution like \( t = ax+b \), where \( dx = \frac{1}{a} dt \).

Question 2. (ii) \(\sqrt{81+(2 x+1)^{2}}\)
Answer:
To integrate this expression, we will first use a substitution.
\[ \int \sqrt{81 + (2x+1)^2} \, dx \]
\[ = \int \sqrt{9^2 + (2x+1)^2} \, dx \]
Let \( t = 2x+1 \). When we take the derivative, we find \( 2 \, dx = dt \), which means that \( dx = \frac{1}{2} \, dt \).
\[ = \int \sqrt{9^2 + t^2} \frac{1}{2} \, dt \]
\[ = \frac{1}{2} \int \sqrt{9^2 + t^2} \, dt \]
We will now use the standard integration formula for \( \int \sqrt{x^2+a^2} \, dx = \frac{x}{2}\sqrt{x^2+a^2} + \frac{a^2}{2} \log|x + \sqrt{x^2+a^2}| + c \). This formula helps solve integrals of the form \( \sqrt{a^2 + x^2} \).
Applying this formula with \( t \) and \( a=9 \):
\[ = \frac{1}{2} \left[ \frac{t}{2}\sqrt{t^2+9^2} + \frac{9^2}{2} \log|t + \sqrt{t^2+9^2}| \right] + c \]
Finally, we substitute \( t \) back with \( 2x+1 \):
\[ = \frac{1}{2} \left[ \frac{2x+1}{2}\sqrt{(2x+1)^2+9^2} + \frac{81}{2} \log|2x+1 + \sqrt{(2x+1)^2+81}| \right] + c \]
\[ = \frac{1}{4} (2x+1)\sqrt{(2x+1)^2+81} + \frac{81}{4} \log|2x+1 + \sqrt{(2x+1)^2+81}| + c \]
In simple words: We simplified the expression using substitution, then applied a well-known integration formula, and finally replaced the substituted term to get the answer.

🎯 Exam Tip: Be careful with the constant factor \( \frac{1}{a} \) arising from the substitution \( t = ax+b \) when integrating. It's easy to miss or misplace.

Question 2. (iii) \(\sqrt{(x+1)^{2}-4}\)
Answer:
To integrate this expression, we first rewrite the constant term to show it as a square.
\[ \int \sqrt{(x+1)^2 - 4} \, dx \]
\[ = \int \sqrt{(x+1)^2 - 2^2} \, dx \]
Let \( t = x+1 \). When we differentiate, we get \( dx = dt \).
\[ = \int \sqrt{t^2 - 2^2} \, dt \]
We use the standard integration formula for \( \int \sqrt{x^2-a^2} \, dx = \frac{x}{2}\sqrt{x^2-a^2} - \frac{a^2}{2} \log|x + \sqrt{x^2-a^2}| + c \). This formula is specifically for integrals involving the square root of a variable squared minus a constant squared.
Applying this formula with \( t \) and \( a=2 \):
\[ = \frac{t}{2}\sqrt{t^2-2^2} - \frac{2^2}{2} \log|t + \sqrt{t^2-2^2}| + c \]
Finally, we substitute \( t \) back with \( x+1 \):
\[ = \frac{x+1}{2}\sqrt{(x+1)^2-2^2} - \frac{4}{2} \log|x+1 + \sqrt{(x+1)^2-4}| + c \]
\[ = \frac{x+1}{2}\sqrt{(x+1)^2-4} - 2 \log|x+1 + \sqrt{(x+1)^2-4}| + c \]
In simple words: We made a simple substitution to change the integral into a standard form. Then, we used a known formula to solve it and put the original terms back to get the final answer.

🎯 Exam Tip: Remember to clearly identify \( x \) and \( a \) in the standard integration formulas to avoid mistakes in substitution.