OP Malhotra Class 12 Maths Solutions Chapter 15 Indefinite Integral 3 Exercise 15 (D)

S Chand Class 12 ICSE Maths Solutions Chapter 15 Indefinite Integral-3 Ex 15(d)

Question 1.
\( \int \frac{d x}{\sqrt{1-x^2}} \)
Answer:
Let \( I = \int \frac{d x}{\sqrt{1-x^2}} \)
We know the standard integral formula: \( \int \frac{d x}{\sqrt{a^2-x^2}} = \sin^{-1} \left(\frac{x}{a}\right) + C \).
Here, \( a^2 = 1 \), so \( a = 1 \).
So, \( I = \sin^{-1} \left(\frac{x}{1}\right) + C \)
\( \implies I = \sin^{-1} x + C \). This integral form is fundamental in trigonometry.
In simple words: This is a direct formula. When you see `dx` over the square root of 1 minus `x` squared, the answer is always inverse sine of `x` plus a constant.

🎯 Exam Tip: Recognizing standard integral forms instantly is crucial for saving time. Memorize the basic formulas, especially those involving inverse trigonometric functions.

Question 2.
\( \int \frac{d x}{\sqrt{16-25 x^2}} \)
Answer:
Let \( I = \int \frac{d x}{\sqrt{16-25 x^2}} \)
First, factor out the coefficient of \( x^2 \) from inside the square root to match the standard form.
\( I = \int \frac{d x}{\sqrt{25 \left(\frac{16}{25}-x^2\right)}} \)
\( \implies I = \int \frac{d x}{5 \sqrt{\left(\frac{4}{5}\right)^2-x^2}} \)
\( \implies I = \frac{1}{5} \int \frac{d x}{\sqrt{\left(\frac{4}{5}\right)^2-x^2}} \)
Now, use the formula \( \int \frac{d x}{\sqrt{a^2-x^2}} = \sin^{-1} \left(\frac{x}{a}\right) + C \), where \( a = \frac{4}{5} \).
\( \implies I = \frac{1}{5} \sin^{-1} \left(\frac{x}{\frac{4}{5}}\right) + C \)
\( \implies I = \frac{1}{5} \sin^{-1} \left(\frac{5x}{4}\right) + C \). Always simplify the fraction in the argument of the inverse function.
In simple words: First, take out the number linked with \( x^2 \) from under the square root. Then, write the remaining number as a square (like \( (\frac{4}{5})^2 \)). Now it matches a common formula for inverse sine.

🎯 Exam Tip: When \( x^2 \) has a coefficient, always factor it out of the square root first to get \( x^2 \) by itself, then adjust the 'a' value accordingly. Don't forget to take the square root of the factored coefficient when bringing it outside the radical.

Question 3.
\( \int \frac{d x}{\sqrt{4 x^2+9}} \)
Answer:
Let \( I = \int \frac{d x}{\sqrt{4 x^2+9}} \)
Factor out the coefficient of \( x^2 \) from the square root.
\( I = \int \frac{d x}{\sqrt{4 \left(x^2+\frac{9}{4}\right)}} \)
\( \implies I = \int \frac{d x}{2 \sqrt{x^2+\left(\frac{3}{2}\right)^2}} \)
\( \implies I = \frac{1}{2} \int \frac{d x}{\sqrt{x^2+\left(\frac{3}{2}\right)^2}} \)
Use the formula \( \int \frac{d x}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right| + C \), where \( a = \frac{3}{2} \).
\( \implies I = \frac{1}{2} \log\left|x+\sqrt{x^2+\left(\frac{3}{2}\right)^2}\right| + C \)
\( \implies I = \frac{1}{2} \log\left|x+\sqrt{x^2+\frac{9}{4}}\right| + C \). This result is related to the hyperbolic sine function.
In simple words: Pull out the number in front of \( x^2 \) from under the square root. Then write the constant part as a square. Now use the special logarithm formula for square roots.

🎯 Exam Tip: Be careful with the signs under the square root. \( x^2+a^2 \) always leads to a logarithm, while \( a^2-x^2 \) leads to inverse sine, and \( x^2-a^2 \) also leads to a logarithm but with a different form.

Question 4.
\( \int \frac{d x}{\sqrt{9 x^2-25}} \)
Answer:
Let \( I = \int \frac{d x}{\sqrt{9 x^2-25}} \)
Factor out the coefficient of \( x^2 \) from the square root.
\( I = \int \frac{d x}{\sqrt{9 \left(x^2-\frac{25}{9}\right)}} \)
\( \implies I = \int \frac{d x}{3 \sqrt{x^2-\left(\frac{5}{3}\right)^2}} \)
\( \implies I = \frac{1}{3} \int \frac{d x}{\sqrt{x^2-\left(\frac{5}{3}\right)^2}} \)
Use the formula \( \int \frac{d x}{\sqrt{x^2-a^2}} = \log\left|x+\sqrt{x^2-a^2}\right| + C \), where \( a = \frac{5}{3} \).
\( \implies I = \frac{1}{3} \log\left|x+\sqrt{x^2-\left(\frac{5}{3}\right)^2}\right| + C \)
\( \implies I = \frac{1}{3} \log\left|x+\sqrt{x^2-\frac{25}{9}}\right| + C \). This type of integral often appears in physics problems.
In simple words: First, take out the number that is with \( x^2 \) from under the square root. Then, write the constant part as a square. Now it fits the logarithm formula for when \( x^2 \) comes before the constant.

🎯 Exam Tip: The formulas for \( \int \frac{dx}{\sqrt{x^2+a^2}} \) and \( \int \frac{dx}{\sqrt{x^2-a^2}} \) are very similar, both involving logarithms. Pay close attention to the sign before \( a^2 \).

Question 5.
\( \int \frac{d x}{\sqrt{a^2-b^2 x^2}} \)
Answer:
Let \( I = \int \frac{d x}{\sqrt{a^2-b^2 x^2}} \)
Factor out \( b^2 \) from inside the square root.
\( I = \int \frac{d x}{\sqrt{b^2 \left(\frac{a^2}{b^2}-x^2\right)}} \)
\( \implies I = \int \frac{d x}{b \sqrt{\left(\frac{a}{b}\right)^2-x^2}} \)
\( \implies I = \frac{1}{b} \int \frac{d x}{\sqrt{\left(\frac{a}{b}\right)^2-x^2}} \)
Use the standard integral formula \( \int \frac{d x}{\sqrt{k^2-x^2}} = \sin^{-1} \left(\frac{x}{k}\right) + C \), where \( k = \frac{a}{b} \).
\( \implies I = \frac{1}{b} \sin^{-1} \left(\frac{x}{\frac{a}{b}}\right) + C \)
\( \implies I = \frac{1}{b} \sin^{-1} \left(\frac{bx}{a}\right) + C \). This generalizes the inverse sine form for varying coefficients.
In simple words: When \( x^2 \) has a number in front of it, take that number out of the square root. This leaves a new constant, which you can use in the inverse sine formula.

🎯 Exam Tip: Generalizing formulas with 'a' and 'b' instead of specific numbers helps understand the pattern. Always remember to extract the square root of the coefficient for \( x^2 \) as `b` and not `b^2` when bringing it outside the radical.

Question 6.
\( \int \frac{d x}{\sqrt{(2-x)^2+1}} \)
Answer:
Let \( I = \int \frac{d x}{\sqrt{(2-x)^2+1}} \)
Use substitution: Let \( t = 2-x \).
Then \( dt = -dx \), which means \( dx = -dt \).
Substitute these into the integral:
\( I = \int \frac{-dt}{\sqrt{t^2+1^2}} \)
\( \implies I = - \int \frac{dt}{\sqrt{t^2+1^2}} \)
Now, use the formula \( \int \frac{d x}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right| + C \), where \( a = 1 \).
\( \implies I = - \log\left|t+\sqrt{t^2+1^2}\right| + C \)
Substitute back \( t = 2-x \):
\( \implies I = - \log\left|(2-x)+\sqrt{(2-x)^2+1}\right| + C \). Handling `(a-x)` terms requires a negative sign from the substitution.
In simple words: When you see `(2-x)` in the integral, replace it with `t`. Remember that `dx` becomes `-dt`. Then, use the logarithm formula for square roots and put `(2-x)` back at the end.

🎯 Exam Tip: For substitution with \( (a-x) \) or \( (x-a) \) forms, remember that \( d(a-x) = -dx \), which introduces a negative sign. Always substitute back the original variable to get the final answer in terms of \( x \).

Question 7.
\( \int \frac{x+2}{\sqrt{x^2+9}} d x \)
Answer:
Let \( I = \int \frac{x+2}{\sqrt{x^2+9}} d x \)
Split the integral into two parts based on the numerator:
\( I = \int \frac{x}{\sqrt{x^2+9}} d x + \int \frac{2}{\sqrt{x^2+9}} d x \)
For the first integral, \( \int \frac{x}{\sqrt{x^2+9}} d x \):
Let \( u = x^2+9 \). Then \( du = 2x \, dx \), so \( x \, dx = \frac{1}{2} du \).
\( \int \frac{1}{\sqrt{u}} \frac{1}{2} du = \frac{1}{2} \int u^{-1/2} du \)
\( = \frac{1}{2} \frac{u^{1/2}}{1/2} = \sqrt{u} = \sqrt{x^2+9} \).
For the second integral, \( \int \frac{2}{\sqrt{x^2+9}} d x \):
\( = 2 \int \frac{d x}{\sqrt{x^2+3^2}} \)
Use the formula \( \int \frac{d x}{\sqrt{x^2+a^2}} = \log\left|x+\sqrt{x^2+a^2}\right| + C \), where \( a = 3 \).
\( = 2 \log\left|x+\sqrt{x^2+3^2}\right| = 2 \log\left|x+\sqrt{x^2+9}\right| \).
Combining both parts:
\( I = \sqrt{x^2+9} + 2 \log\left|x+\sqrt{x^2+9}\right| + C \). Splitting the fraction makes complex integrals manageable.
In simple words: Break the question into two parts. For the first part, if the top is related to the derivative of the bottom, use substitution. For the second part, use the standard logarithm formula. Then add both answers together.

🎯 Exam Tip: When the numerator contains both a variable term and a constant, split the integral. One part often becomes a simple substitution, while the other becomes a standard formula.

Question 8.
\( \int \frac{e^x d x}{\sqrt{4-e^{2 x}}} \)
Answer:
Let \( I = \int \frac{e^x d x}{\sqrt{4-e^{2 x}}} \)
Rewrite \( e^{2x} \) as \( (e^x)^2 \).
\( I = \int \frac{e^x d x}{\sqrt{4-(e^x)^2}} \)
Use substitution: Let \( t = e^x \).
Then \( dt = e^x \, dx \).
Substitute these into the integral:
\( I = \int \frac{d t}{\sqrt{4-t^2}} \)
\( \implies I = \int \frac{d t}{\sqrt{2^2-t^2}} \)
Now, use the formula \( \int \frac{d x}{\sqrt{a^2-x^2}} = \sin^{-1} \left(\frac{x}{a}\right) + C \), where \( a = 2 \).
\( \implies I = \sin^{-1} \left(\frac{t}{2}\right) + C \)
Substitute back \( t = e^x \):
\( \implies I = \sin^{-1} \left(\frac{e^x}{2}\right) + C \). This shows how exponential functions can simplify into inverse trigonometric forms.
In simple words: Notice that \( e^x \) is the derivative of \( e^x \). So, let \( t \) be \( e^x \). The integral then turns into a simple inverse sine form.

🎯 Exam Tip: Always look for relationships between parts of the integrand. Here, \( e^x \) is the derivative of \( e^x \), which signals a substitution opportunity. Also, recognize \( e^{2x} = (e^x)^2 \).

Question 9.
\( \int \frac{x^2}{\sqrt{x^6-a^6}} d x \)
Answer:
Let \( I = \int \frac{x^2}{\sqrt{x^6-a^6}} d x \)
Rewrite \( x^6 \) as \( (x^3)^2 \) and \( a^6 \) as \( (a^3)^2 \).
\( I = \int \frac{x^2}{\sqrt{(x^3)^2-(a^3)^2}} d x \)
Use substitution: Let \( t = x^3 \).
Then \( dt = 3x^2 \, dx \), so \( x^2 \, dx = \frac{1}{3} dt \).
Substitute these into the integral:
\( I = \int \frac{\frac{1}{3} dt}{\sqrt{t^2-(a^3)^2}} \)
\( \implies I = \frac{1}{3} \int \frac{d t}{\sqrt{t^2-(a^3)^2}} \)
Now, use the formula \( \int \frac{d x}{\sqrt{x^2-k^2}} = \log\left|x+\sqrt{x^2-k^2}\right| + C \), where \( k = a^3 \).
\( \implies I = \frac{1}{3} \log\left|t+\sqrt{t^2-(a^3)^2}\right| + C \)
Substitute back \( t = x^3 \):
\( \implies I = \frac{1}{3} \log\left|x^3+\sqrt{(x^3)^2-(a^3)^2}\right| + C \)
\( \implies I = \frac{1}{3} \log\left|x^3+\sqrt{x^6-a^6}\right| + C \). Powers of variables can often be simplified through substitution.
In simple words: Notice that \( x^2 \) is part of the derivative of \( x^3 \). So, let \( t \) be \( x^3 \). This makes the integral look like a standard logarithm form.

🎯 Exam Tip: When dealing with higher powers like \( x^6 \), try to express them as squares of a lower power (e.g., \( (x^3)^2 \)) to match standard integral forms, then use substitution.

Question 10.
\( \int \frac{\sec ^2 x}{\tan ^2 x+4} d x \)
Answer:
Let \( I = \int \frac{\sec ^2 x}{\tan ^2 x+4} d x \)
Use substitution: Let \( t = \tan x \).
Then \( dt = \sec^2 x \, dx \).
Substitute these into the integral:
\( I = \int \frac{d t}{t^2+4} \)
\( \implies I = \int \frac{d t}{t^2+2^2} \)
Now, use the formula \( \int \frac{d x}{x^2+a^2} = \frac{1}{a} \tan^{-1} \left(\frac{x}{a}\right) + C \), where \( a = 2 \).
\( \implies I = \frac{1}{2} \tan^{-1} \left(\frac{t}{2}\right) + C \)
Substitute back \( t = \tan x \):
\( \implies I = \frac{1}{2} \tan^{-1} \left(\frac{\tan x}{2}\right) + C \). Trigonometric substitutions are common in integration problems.
In simple words: See that \( \sec^2 x \) is the derivative of \( \tan x \). So, replace \( \tan x \) with \( t \). The integral then becomes a simple inverse tangent formula.

🎯 Exam Tip: Recognizing that the numerator is the derivative of a part of the denominator (or its base function) is key to applying substitution effectively. Here, \( \sec^2 x \) is the derivative of \( \tan x \).

Question 11.
\( \int \frac{a^x}{\sqrt{1-a^{2 x}}} d x \)
Answer:
Let \( I = \int \frac{a^x}{\sqrt{1-a^{2 x}}} d x \)
Rewrite \( a^{2x} \) as \( (a^x)^2 \).
\( I = \int \frac{a^x}{\sqrt{1-(a^x)^2}} d x \)
Use substitution: Let \( t = a^x \).
Then \( dt = a^x \log a \, dx \), so \( a^x \, dx = \frac{1}{\log a} dt \).
Substitute these into the integral:
\( I = \int \frac{\frac{1}{\log a} dt}{\sqrt{1-t^2}} \)
\( \implies I = \frac{1}{\log a} \int \frac{d t}{\sqrt{1^2-t^2}} \)
Now, use the formula \( \int \frac{d x}{\sqrt{k^2-x^2}} = \sin^{-1} \left(\frac{x}{k}\right) + C \), where \( k = 1 \).
\( \implies I = \frac{1}{\log a} \sin^{-1} \left(\frac{t}{1}\right) + C \)
Substitute back \( t = a^x \):
\( \implies I = \frac{1}{\log a} \sin^{-1} (a^x) + C \). This demonstrates an inverse sine integral involving an exponential base other than `e`.
In simple words: Let \( t \) be \( a^x \). Remember that the derivative of \( a^x \) is \( a^x \log a \). This substitution helps transform the integral into a basic inverse sine form. Don't forget the \( \frac{1}{\log a} \) factor.

🎯 Exam Tip: When substituting with \( a^x \), the derivative includes \( \log a \). Make sure to account for this constant factor when rewriting the integral in terms of \( dt \).

Question 12.
\( \int \sqrt{\frac{1-x}{1+x}} d x \)
Answer:
Let \( I = \int \sqrt{\frac{1-x}{1+x}} d x \)
To simplify, rationalize the integrand by multiplying the numerator and denominator inside the square root by \( (1-x) \).
\( I = \int \sqrt{\frac{1-x}{1+x} \times \frac{1-x}{1-x}} d x \)
\( \implies I = \int \sqrt{\frac{(1-x)^2}{1-x^2}} d x \)
\( \implies I = \int \frac{1-x}{\sqrt{1-x^2}} d x \)
Now, split the integral into two parts:
\( I = \int \frac{1}{\sqrt{1-x^2}} d x - \int \frac{x}{\sqrt{1-x^2}} d x \)
For the first integral, \( \int \frac{1}{\sqrt{1-x^2}} d x \):
This is a standard formula: \( = \sin^{-1} x \).
For the second integral, \( \int \frac{x}{\sqrt{1-x^2}} d x \):
Let \( u = 1-x^2 \). Then \( du = -2x \, dx \), so \( x \, dx = -\frac{1}{2} du \).
\( \int \frac{1}{\sqrt{u}} \left(-\frac{1}{2}\right) du = -\frac{1}{2} \int u^{-1/2} du \)
\( = -\frac{1}{2} \frac{u^{1/2}}{1/2} = - \sqrt{u} = -\sqrt{1-x^2} \).
Combining both results:
\( I = \sin^{-1} x - (-\sqrt{1-x^2}) + C \)
\( \implies I = \sin^{-1} x + \sqrt{1-x^2} + C \). Rationalization often helps reveal standard integral forms.
In simple words: Multiply the top and bottom inside the square root by `(1-x)` to get `(1-x)` on top and `sqrt(1-x^2)` on the bottom. Then, split it into two simple integrals: one for inverse sine and one for a square root via substitution.

🎯 Exam Tip: For rational functions under a square root, rationalizing the numerator or denominator can transform the integral into a solvable form, often leading to a sum or difference of standard integrals.

Question 13.
\( \int \frac{(x-1)}{\sqrt{x^2-1}} d x \)
Answer:
Let \( I = \int \frac{(x-1)}{\sqrt{x^2-1}} d x \)
Split the integral into two parts based on the numerator:
\( I = \int \frac{x}{\sqrt{x^2-1}} d x - \int \frac{1}{\sqrt{x^2-1}} d x \)
For the first integral, \( \int \frac{x}{\sqrt{x^2-1}} d x \):
Let \( u = x^2-1 \). Then \( du = 2x \, dx \), so \( x \, dx = \frac{1}{2} du \).
\( \int \frac{1}{\sqrt{u}} \frac{1}{2} du = \frac{1}{2} \int u^{-1/2} du \)
\( = \frac{1}{2} \frac{u^{1/2}}{1/2} = \sqrt{u} = \sqrt{x^2-1} \).
For the second integral, \( \int \frac{1}{\sqrt{x^2-1}} d x \):
This is a standard formula: \( = \log\left|x+\sqrt{x^2-1}\right| \).
Combining both parts:
\( I = \sqrt{x^2-1} - \log\left|x+\sqrt{x^2-1}\right| + C \). Splitting the fraction simplifies complex integrands.
In simple words: Break this integral into two pieces. One part can be solved by letting \( u \) be \( x^2-1 \). The other part is a direct formula involving a logarithm. Put the two answers together for the final solution.

🎯 Exam Tip: Similar to Question 7, splitting the numerator into multiple terms often allows you to apply a combination of substitution and standard integral formulas. Always check for direct derivative relationships.