RBSE Solutions Class 12 Maths Chapter 9 Integration Exercise 9.2

Question 1. Integrate the following w.r.t x:
(a) x sin x²
(b) x √ x² + 1
Answer:
(a) To integrate \( x \sin x^2 \), we use a substitution method. Let \( t = x^2 \), which means \( dt = 2x \, dx \). So, \( x \, dx \) becomes \( \frac{dt}{2} \). The integral changes to \( \frac{1}{2} \int \sin t \, dt \). The integral of \( \sin t \) is \( -\cos t \). Therefore, the result is \( -\frac{1}{2} \cos x^2 + C \). This technique simplifies complex integrals by changing variables.
(b) For integrating \( x \sqrt{x^2+1} \), we substitute \( t = x^2+1 \). Differentiating both sides gives \( dt = 2x \, dx \), so \( x \, dx = \frac{dt}{2} \). The integral transforms into \( \frac{1}{2} \int \sqrt{t} \, dt \), which is \( \frac{1}{2} \int t^{1/2} \, dt \). Integrating \( t^{1/2} \) gives \( \frac{t^{3/2}}{3/2} \). After simplifying and substituting back \( t = x^2+1 \), the final result is \( \frac{1}{3} (x^2+1)^{3/2} + C \). This method is very useful when one part of the function is the derivative of another part.
In simple words: For part (a), we let \( x^2 \) be \( t \), changing \( x \, dx \) to \( \frac{dt}{2} \), which simplifies the integral. For part (b), we let \( x^2+1 \) be \( t \), changing \( x \, dx \) to \( \frac{dt}{2} \), making it easier to solve. We always add \( C \) for indefinite integrals.

🎯 Exam Tip: Always look for a suitable substitution where one part of the integrand is the derivative of another part. Remember to change the differential \( dx \) to \( dt \) after substitution.

Question 2. Find the following:
(a) \( \int \frac{e^x - \sin x}{e^x + \cos x} dx \)
(b) \( \int \frac{e^x}{\sqrt{1+e^x}} dx \)
Answer:
(a) To solve \( \int \frac{e^x - \sin x}{e^x + \cos x} dx \), we notice that the numerator is the derivative of the denominator. So, we let \( t = e^x + \cos x \). This makes \( dt = (e^x - \sin x) dx \). The integral simplifies to \( \int \frac{dt}{t} \), which integrates to \( \log |t| + C \). Replacing \( t \) with \( e^x + \cos x \), we get the final answer: \( \log |e^x + \cos x| + C \). This is a standard method for integrals of the form \( \int \frac{f'(x)}{f(x)} dx \).
(b) For \( \int \frac{e^x}{\sqrt{1+e^x}} dx \), we use substitution by setting \( t = 1+e^x \). When we differentiate \( t \), we get \( dt = e^x dx \). The integral then changes to \( \int \frac{dt}{\sqrt{t}} \), which is \( \int t^{-1/2} dt \). Integrating \( t^{-1/2} \) gives \( \frac{t^{1/2}}{1/2} \), or \( 2\sqrt{t} \). Finally, substituting \( t \) back with \( 1+e^x \), the solution is \( 2\sqrt{1+e^x} + C \). This substitution technique is powerful for square root expressions.
In simple words: For part (a), since the top part is the derivative of the bottom part, the answer is the logarithm of the bottom part. For part (b), we let \( 1+e^x \) be \( t \), which makes \( e^x dx \) into \( dt \), simplifying the integral. Always remember the constant \( C \).

🎯 Exam Tip: For rational functions, always check if the numerator is the derivative of the denominator. For square root functions, substituting the expression inside the root often works well.

Question 3. Find the following:
(a) \( \int \sqrt{e^x+1} dx \)
(b) \( \int \frac{e^{\sqrt{x}} \cos(e^{\sqrt{x}})}{\sqrt{x}} dx \)
Answer:
(a) To integrate \( \sqrt{e^x+1} \), we use the substitution \( y = \sqrt{e^x+1} \). This means \( e^x+1 = y^2 \), and \( dx = \frac{2y}{y^2-1} dy \). The integral becomes \( 2 \int \frac{y^2}{y^2-1} dy \). We rewrite \( \frac{y^2}{y^2-1} \) as \( 1 + \frac{1}{y^2-1} \). Then we integrate each part separately. The integral of \( \frac{1}{y^2-1} \) is \( \frac{1}{2} \log \left|\frac{y-1}{y+1}\right| \). Substituting \( y \) back with \( \sqrt{e^x+1} \), we get the final answer: \( 2\sqrt{e^x+1} + \log \left|\frac{\sqrt{e^x+1}-1}{\sqrt{e^x+1}+1}\right| + C \). This method helps convert complicated square root integrals into rational functions.
(b) For \( \int \frac{e^{\sqrt{x}} \cos(e^{\sqrt{x}})}{\sqrt{x}} dx \), we let \( t = e^{\sqrt{x}} \). When we differentiate \( t \), we get \( dt = e^{\sqrt{x}} \cdot \frac{1}{2\sqrt{x}} dx \), which means \( \frac{e^{\sqrt{x}}}{\sqrt{x}} dx = 2 dt \). The integral simplifies to \( \int \cos t \cdot 2 dt \), or \( 2 \int \cos t dt \). The integral of \( \cos t \) is \( \sin t \). So, the answer is \( 2 \sin t + C \). Substituting back \( t = e^{\sqrt{x}} \), the final solution is \( 2 \sin(e^{\sqrt{x}}) + C \). This substitution is effective when the integrand contains a function and its derivative.
In simple words: For part (a), we substitute \( y \) for \( \sqrt{e^x+1} \) to make the integral a fraction that can be split and integrated. For part (b), we let \( t = e^{\sqrt{x}} \), which helps to transform the integral into a simpler cosine integral. Always add \( C \) at the end.

🎯 Exam Tip: When dealing with integrals involving \( \sqrt{e^x+k} \), consider a full substitution of the square root itself. For exponential functions in the argument of a trig function, try substituting the exponential term.

Question 4. Find the following:
(a) \( \int \frac{1}{x(1 + \log x)} dx \)
(b) \( \int \frac{(1 + \log x)^3}{x} dx \)
Answer:
(a) To integrate \( \frac{1}{x(1+\log x)} \), we use substitution. Let \( t = 1+\log x \). Then, its derivative, \( dt = \frac{1}{x} dx \). The integral transforms into a simpler form: \( \int \frac{1}{t} dt \). This integrates to \( \log |t| + C \). Substituting \( t \) back with \( 1+\log x \), the final result is \( \log |1+\log x| + C \). This is a common pattern where a function and its derivative are present in the integrand.
(b) For the integral \( \int \frac{(1+\log x)^3}{x} dx \), we perform a substitution. Let \( t = 1+\log x \). Then, \( dt = \frac{1}{x} dx \). The integral simplifies considerably to \( \int t^3 dt \). Integrating \( t^3 \) gives \( \frac{t^4}{4} \). After replacing \( t \) with \( 1+\log x \), the solution is \( \frac{(1+\log x)^4}{4} + C \). This technique is very useful for power functions within an integral.
In simple words: For both parts (a) and (b), we let \( 1+\log x \) be \( t \). Then, \( \frac{1}{x} dx \) becomes \( dt \). This makes both integrals much simpler to solve. After solving, we put \( 1+\log x \) back in place of \( t \). Always add the constant \( C \).

🎯 Exam Tip: In integrals involving \( \log x \), consider substituting the logarithmic term. Its derivative \( \frac{1}{x} \) often helps simplify the integral.

Question 5. Find the following:
(a) \( \int \frac{e^{m \tan^{-1} x}}{1+x^2} dx \)
(b) \( \int \frac{\sin^P x}{\cos^P x \cdot \cos^2 x} dx \)
Answer:
(a) To integrate \( \frac{e^{m \tan^{-1} x}}{1+x^2} dx \), we let \( t = m \tan^{-1} x \). Upon differentiating, we get \( dt = m \cdot \frac{1}{1+x^2} dx \), so \( \frac{1}{1+x^2} dx = \frac{dt}{m} \). The integral then simplifies to \( \int e^t \frac{dt}{m} \), which is \( \frac{1}{m} \int e^t dt \). The integral of \( e^t \) is \( e^t \). Thus, the result is \( \frac{1}{m} e^{m \tan^{-1} x} + C \). This substitution is effective for exponents involving inverse trigonometric functions.
(b) For \( \int \frac{\sin^P x}{\cos^P x \cdot \cos^2 x} dx \), we first rewrite the integrand as \( \int \tan^P x \cdot \sec^2 x \, dx \). Now, we use the substitution \( t = \tan x \). Differentiating gives \( dt = \sec^2 x dx \). The integral becomes \( \int t^P dt \). Assuming \( P \neq -1 \), this integrates to \( \frac{t^{P+1}}{P+1} \). Finally, substituting \( t \) back with \( \tan x \), the solution is \( \frac{(\tan x)^{P+1}}{P+1} + C \). This method is powerful when a function and its derivative are present in a multiplicative form.
In simple words: For part (a), we substitute the exponent \( m \tan^{-1} x \) with \( t \) to simplify the integral to \( \frac{1}{m} \int e^t dt \). For part (b), we rewrite the integral using \( \tan x \) and \( \sec^2 x \), then let \( t = \tan x \), which changes it into a simple power integral \( \int t^P dt \). Add \( C \) at the end.

🎯 Exam Tip: For exponential functions with complex exponents, consider substituting the entire exponent. For trigonometric integrals, rewriting expressions using identities like \( \tan x = \frac{\sin x}{\cos x} \) and \( \sec^2 x = 1+\tan^2 x \) can reveal easier substitutions.

Question 6. Find the following:
(a) \( \int \frac{1}{\sqrt{1+\cos 2x}} dx \)
(b) \( \int \frac{1+\cos x}{\sin x \cos x} dx \)
Answer:
(a) To integrate \( \frac{1}{\sqrt{1+\cos 2x}} dx \), we use the identity \( 1+\cos 2x = 2\cos^2 x \). So the integral becomes \( \int \frac{1}{\sqrt{2\cos^2 x}} dx = \int \frac{1}{\sqrt{2} |\cos x|} dx \). Assuming \( \cos x > 0 \), this simplifies to \( \frac{1}{\sqrt{2}} \int \sec x dx \). The integral of \( \sec x \) is \( \log |\sec x + \tan x| \). Thus, the final answer is \( \frac{1}{\sqrt{2}} \log |\sec x + \tan x| + C \). Trigonometric identities are key for simplifying such integrals.
(b) For \( \int \frac{1+\cos x}{\sin x \cos x} dx \), we split the fraction into two parts: \( \int \frac{1}{\sin x \cos x} dx + \int \frac{\cos x}{\sin x \cos x} dx \). The second part simplifies to \( \int \frac{1}{\sin x} dx = \int \operatorname{cosec} x dx \). For the first part, we can rewrite it as \( 2 \int \frac{1}{2\sin x \cos x} dx = 2 \int \frac{1}{\sin 2x} dx = 2 \int \operatorname{cosec} 2x dx \). The integral of \( \operatorname{cosec}(ax) \) is \( \frac{1}{a} \log |\operatorname{cosec}(ax) - \cot(ax)| \). After integrating both parts and adding them, the solution is \( \log |\operatorname{cosec} 2x - \cot 2x| + \log |\operatorname{cosec} x - \cot x| + C \). Splitting fractions often simplifies complex integrands.
In simple words: For part (a), we change \( 1+\cos 2x \) to \( 2\cos^2 x \) and then take the square root. We integrate \( \sec x \). For part (b), we split the fraction into two parts. The first part is changed to \( 2 \operatorname{cosec} 2x \), and the second part is \( \operatorname{cosec} x \). We then integrate both using standard formulas. Always add \( C \).

🎯 Exam Tip: Remember double angle identities like \( 1+\cos 2x = 2\cos^2 x \) and \( \sin 2x = 2\sin x \cos x \). When splitting fractions, simplify terms before integrating. Be careful with absolute values for logarithmic terms.

Question 7. Find the following:
(a) \( \int \sin 3x \sin 2x dx \)
(b) \( \int \sqrt{1 - \sin x} dx \)
Answer:
(a) To solve \( \int \sin 3x \sin 2x dx \), we use the trigonometric identity \( 2 \sin A \sin B = \cos(A-B) - \cos(A+B) \). So, \( \sin 3x \sin 2x \) becomes \( \frac{1}{2} (\cos x - \cos 5x) \). We then integrate each cosine term separately. The integral of \( \cos x \) is \( \sin x \), and the integral of \( \cos 5x \) is \( \frac{\sin 5x}{5} \). Combining these, the final answer is \( \frac{1}{2} \left( \sin x - \frac{\sin 5x}{5} \right) + C \). Product-to-sum identities are crucial for integrating products of sine and cosine functions.
(b) For \( \int \sqrt{1 - \sin x} dx \), we transform the expression inside the square root using half-angle identities: \( 1 = \sin^2 \frac{x}{2} + \cos^2 \frac{x}{2} \) and \( \sin x = 2 \sin \frac{x}{2} \cos \frac{x}{2} \). This turns the integrand into \( \sqrt{\left(\cos \frac{x}{2} - \sin \frac{x}{2}\right)^2} \), which simplifies to \( \left|\cos \frac{x}{2} - \sin \frac{x}{2}\right| \). The solution evaluates this to \( \pm \left(\cos \frac{x}{2} + \sin \frac{x}{2}\right) \). The integral is \( \pm 2 \left(\sin \frac{x}{2} + \cos \frac{x}{2}\right) + C \). This technique of converting to squares under a root is powerful in trigonometry.
In simple words: For part (a), we use a special formula to change the product of sines into a difference of cosines, making it easy to integrate. For part (b), we use half-angle formulas to change \( 1 - \sin x \) into a perfect square under the root. Then we take the square root and integrate. Always add \( C \) at the end.

🎯 Exam Tip: When integrating products of trigonometric functions, remember product-to-sum identities. For expressions under square roots involving \( 1 \pm \sin x \) or \( 1 \pm \cos x \), think of half-angle identities to create a perfect square.

Question 8. Find the following:
(a) \( \int \cos^4 x dx \)
(b) \( \int \sin^3 x dx \)
Answer:
(a) To integrate \( \cos^4 x \), we rewrite it as \( (\cos^2 x)^2 \). Using the identity \( \cos^2 x = \frac{1+\cos 2x}{2} \), the integral becomes \( \frac{1}{4} \int (1+2\cos 2x+\cos^2 2x) dx \). We apply the identity again for \( \cos^2 2x = \frac{1+\cos 4x}{2} \). After simplifying the terms, we get \( \frac{1}{8} \int (3+4\cos 2x+\cos 4x) dx \). Integrating each term, the result is \( \frac{1}{8} \left( 3x + 2\sin 2x + \frac{\sin 4x}{4} \right) + C \). Reducing powers using half-angle formulas is key for these integrals.
(b) For \( \int \sin^3 x dx \), we use the triple-angle identity for sine: \( \sin 3x = 3\sin x - 4\sin^3 x \). From this, we can express \( \sin^3 x \) as \( \frac{1}{4}(3\sin x - \sin 3x) \). The integral becomes \( \frac{1}{4} \int (3\sin x - \sin 3x) dx \). We then integrate \( \sin x \) and \( \sin 3x \) separately. The integral of \( \sin x \) is \( -\cos x \), and for \( \sin 3x \) is \( -\frac{\cos 3x}{3} \). Combining these, the solution is \( -\frac{3}{4}\cos x + \frac{1}{12}\cos 3x + C \). Using triple-angle formulas simplifies cubic trigonometric integrals.
In simple words: For part (a), we reduce the power of \( \cos^4 x \) by using the double angle formula for \( \cos^2 x \) two times. This breaks it down into simpler cosine terms that we can integrate. For part (b), we use a formula that relates \( \sin^3 x \) to \( \sin x \) and \( \sin 3x \). This allows us to integrate the sine terms directly. Don't forget \( C \).

🎯 Exam Tip: When integrating higher powers of sine or cosine, always try to reduce the power using identities like \( \sin^2 x = \frac{1-\cos 2x}{2} \) and \( \cos^2 x = \frac{1+\cos 2x}{2} \). For odd powers, sometimes identities like \( \sin^3 x = \sin x (1-\cos^2 x) \) also work with substitution.

Question 9. Find the following:
(a) \( \int (\tan x + \frac{1}{\tan x}) \sec^2 x dx \)
(b) \( \int \frac{(1+x)e^x}{\cos^2(xe^x)} dx \)
Answer:
(a) To integrate \( \int (\tan x + \cot x) \sec^2 x dx \), we use a substitution. Let \( t = \tan x \). Then, its derivative, \( dt = \sec^2 x dx \). The term \( \cot x \) can be written as \( \frac{1}{\tan x} \), so it becomes \( \frac{1}{t} \). The integral simplifies to \( \int \left(t + \frac{1}{t}\right) dt \). Integrating term by term, we get \( \frac{t^2}{2} + \log |t| \). Finally, substituting back \( t = \tan x \), the solution is \( \frac{\tan^2 x}{2} + \log |\tan x| + C \). Recognizing parts of the integrand as a function and its derivative is crucial here.
(b) For the integral \( \int \frac{(1+x)e^x}{\cos^2(xe^x)} dx \), we identify \( xe^x \) as a candidate for substitution. Let \( t = xe^x \). Using the product rule to differentiate, \( dt = (1 \cdot e^x + x \cdot e^x) dx = e^x(1+x) dx \). The integral then becomes \( \int \frac{dt}{\cos^2 t} \), which is \( \int \sec^2 t dt \). The integral of \( \sec^2 t \) is \( \tan t \). Substituting back \( t = xe^x \), the final answer is \( \tan(xe^x) + C \). This substitution is very effective when the derivative of an inner function is present in the integrand.
In simple words: For part (a), we let \( t = \tan x \), so \( \sec^2 x dx \) becomes \( dt \), turning the integral into \( \int (t + \frac{1}{t}) dt \). For part (b), we let \( t = xe^x \). Its derivative \( (1+x)e^x dx \) is also in the integral, simplifying it to \( \int \frac{1}{\cos^2 t} dt \), which is \( \int \sec^2 t dt \). Remember to replace \( t \) and add \( C \).

🎯 Exam Tip: When you see a function and its derivative within an integral, consider substitution. For composite functions, try substituting the "inner" function. Pay attention to product rule derivatives for terms like \( xe^x \).

Question 10. Find the following:
(a) \( \int \frac{1}{1 - \tan x} dx \)
(b) \( \int \frac{1}{1 + \cot x} dx \)
Answer:
(a) To integrate \( \frac{1}{1 - \tan x} dx \), we first rewrite \( \tan x \) as \( \frac{\sin x}{\cos x} \) and combine the terms to get \( \int \frac{\cos x}{\cos x - \sin x} dx \). We then use a trick: multiply the numerator by 2 and add/subtract \( \sin x \) to create two parts. This gives \( \frac{1}{2} \int \left(1 + \frac{\cos x + \sin x}{\cos x - \sin x}\right) dx \). The first part integrates to \( \frac{1}{2} x \). For the second part, we notice that the numerator \( \cos x + \sin x \) is almost the negative derivative of the denominator \( \cos x - \sin x \). Using substitution \( t = \cos x - \sin x \), we get \( -\frac{1}{2} \log |t| \). Combining, the final answer is \( \frac{1}{2} x - \frac{1}{2} \log |\cos x - \sin x| + C \). This specific manipulation is key for integrals of this form.
(b) For \( \int \frac{1}{1 + \cot x} dx \), we convert \( \cot x \) to \( \frac{\cos x}{\sin x} \) and simplify to get \( \int \frac{\sin x}{\sin x + \cos x} dx \). Similar to the previous problem, we multiply the numerator by 2 and add/subtract \( \cos x \) to strategically split the integrand. This leads to \( \frac{1}{2} \int \left(1 + \frac{\sin x - \cos x}{\sin x + \cos x}\right) dx \). The integral of the first part is \( \frac{1}{2} x \). For the second part, by substituting \( t = \sin x + \cos x \), we find that \( \sin x - \cos x \) is related to \( dt \). This results in \( -\frac{1}{2} \log |\sin x + \cos x| \). Combining both parts, the solution is \( \frac{1}{2} x - \frac{1}{2} \log |\sin x + \cos x| + C \). These reciprocal trigonometric functions often require algebraic manipulation before integration.
In simple words: For both parts, first change \( \tan x \) or \( \cot x \) to sine and cosine. Then, use a special trick where you modify the numerator to create two simpler parts. One part integrates to \( \frac{1}{2}x \), and the other part uses substitution to become a logarithm. Combine these parts and add \( C \).

🎯 Exam Tip: Integrals of the form \( \int \frac{1}{1 \pm \tan x} dx \) or \( \int \frac{1}{1 \pm \cot x} dx \) are solved by converting to sine and cosine, and then adjusting the numerator to be a sum/difference of the denominator and its derivative. This is a common pattern to memorize.

Question 11. Find the following:
(a) \( \int \frac{\sec^4 x}{\sqrt{\tan x}} dx \)
(b) \( \int \frac{1 - \tan x}{1 + \tan x} dx \)
Answer:
(a) To integrate \( \frac{\sec^4 x}{\sqrt{\tan x}} dx \), we first rewrite \( \sec^4 x \) as \( \sec^2 x \cdot \sec^2 x \). Then, we use the identity \( \sec^2 x = 1+\tan^2 x \). The integral becomes \( \int \frac{(1+\tan^2 x) \sec^2 x}{\sqrt{\tan x}} dx \). We use substitution: let \( t = \tan x \), so \( dt = \sec^2 x dx \). The integral simplifies to \( \int \frac{1+t^2}{\sqrt{t}} dt = \int (t^{-1/2} + t^{3/2}) dt \). Integrating each power term, we get \( 2t^{1/2} + \frac{2}{5} t^{5/2} \). Substituting \( t \) back with \( \tan x \), the solution is \( 2\sqrt{\tan x} + \frac{2}{5} (\tan x)^{5/2} + C \). This technique effectively converts trigonometric powers into algebraic powers.
(b) For \( \int \frac{1 - \tan x}{1 + \tan x} dx \), we first convert \( \tan x \) to \( \frac{\sin x}{\cos x} \) and simplify the expression to \( \int \frac{\cos x - \sin x}{\cos x + \sin x} dx \). We can then use substitution. Let \( t = \cos x + \sin x \). Differentiating \( t \), we get \( dt = (-\sin x + \cos x) dx = (\cos x - \sin x) dx \). This is exactly the numerator. The integral simplifies to \( \int \frac{dt}{t} \), which integrates to \( \log |t| \). Substituting \( t \) back with \( \cos x + \sin x \), the final answer is \( \log |\cos x + \sin x| + C \). This form is an instance of \( \int \frac{f'(x)}{f(x)} dx \).
In simple words: For part (a), we break \( \sec^4 x \) into \( \sec^2 x \cdot \sec^2 x \), change one \( \sec^2 x \) to \( 1+\tan^2 x \), and then substitute \( t=\tan x \). This makes it a simple power integral. For part (b), we change \( \tan x \) to \( \frac{\sin x}{\cos x} \), then notice that the numerator is the derivative of the denominator (after substitution). We then integrate to get a logarithm. Add \( C \) for both.

🎯 Exam Tip: When dealing with powers of \( \sec x \) and \( \tan x \), remember the identity \( \sec^2 x = 1+\tan^2 x \). For rational functions of \( \tan x \), converting to sine and cosine often reveals a substitution opportunity.

Question 12. Find the following:
(a) \( \int \frac{\sin (x-a)}{\sin (x+a)} dx \)
(b) \( \int \frac{\sin x}{\sin (x-a)} dx \)
Answer:
(a) To integrate \( \int \frac{\sin (x-a)}{\sin (x+a)} dx \), we use the substitution \( t = x+a \). This means \( x = t-a \) and \( dx = dt \). The numerator \( \sin(x-a) \) becomes \( \sin(t-2a) \). We then use the trigonometric identity \( \sin(A-B) = \sin A \cos B - \cos A \sin B \) to expand \( \sin(t-2a) \). The integral separates into \( \int \cos 2a \, dt - \int \cot t \sin 2a \, dt \). Since \( a \) is a constant, \( \cos 2a \) and \( \sin 2a \) are constants. We integrate these terms. Finally, substitute \( t \) back with \( x+a \), combine constants, and the solution is \( x \cos 2a - \sin 2a \log |\sin(x+a)| + C \). Manipulating the angles is crucial in these types of integrals.
(b) For \( \int \frac{\sin x}{\sin (x-a)} dx \), we perform a substitution by letting \( t = x-a \). This implies \( x = t+a \) and \( dx = dt \). The numerator \( \sin x \) becomes \( \sin(t+a) \). Using the identity \( \sin(A+B) = \sin A \cos B + \cos A \sin B \), we expand \( \sin(t+a) \). The integral simplifies to \( \int \cos a \, dt + \int \cot t \sin a \, dt \). We treat \( \cos a \) and \( \sin a \) as constants. Integrating gives \( t \cos a + \sin a \log |\sin t| \). Substituting \( t \) back with \( x-a \) and collecting constant terms, the final result is \( x \cos a + \sin a \log |\sin(x-a)| + C \). Strategic substitution is very effective for simplifying integrands with angle differences.
In simple words: For part (a), we substitute \( t = x+a \) and for part (b), we substitute \( t = x-a \). In both cases, this changes the numerator into an expanded sine term using addition/subtraction formulas. We then split the integral and integrate the constant and cotangent terms. Finally, we replace \( t \) and group all constants into \( C \).

🎯 Exam Tip: When the integrand involves \( \sin(x \pm a) \) in the denominator, a substitution of the denominator's angle is often effective. Then use angle sum/difference formulas to expand the numerator, allowing the integral to be split into simpler terms.

Question 13. Find the following:
(a) \( \int \frac{\sin 2x}{\sin 5x \sin 3x} dx \)
(b) \( \int \frac{\sin 2x}{\sin (x-\frac{\pi}{6})\sin (x+\frac{\pi}{6})} dx \)
Answer:
(a) We need to integrate \( I = \int \frac{\sin 2x}{\sin 5x \sin 3x} dx \).
We know that \( \sin 2x = \sin (5x - 3x) \).
Using the identity \( \sin(A-B) = \sin A \cos B - \cos A \sin B \), we get:
\( \sin (5x - 3x) = \sin 5x \cos 3x - \cos 5x \sin 3x \).
So, the integral becomes:
\( I = \int \frac{\sin 5x \cos 3x - \cos 5x \sin 3x}{\sin 5x \sin 3x} dx \)
Now, we can split this into two fractions:
\( = \int \left( \frac{\sin 5x \cos 3x}{\sin 5x \sin 3x} - \frac{\cos 5x \sin 3x}{\sin 5x \sin 3x} \right) dx \)
\( = \int (\cot 3x - \cot 5x) dx \)
We integrate each term separately:
\( = \int \cot 3x dx - \int \cot 5x dx \)
Using the formula \( \int \cot(ax) dx = \frac{1}{a} \log |\sin(ax)| + C \):
\( = \frac{1}{3} \log |\sin 3x| - \frac{1}{5} \log |\sin 5x| + C \)

(b) We need to integrate \( I = \int \frac{\sin 2x}{\sin (x-\frac{\pi}{6})\sin (x+\frac{\pi}{6})} dx \).
We can write \( 2x \) as the sum of the angles in the denominator: \( 2x = (x+\frac{\pi}{6}) + (x-\frac{\pi}{6}) \).
Using the identity \( \sin(A+B) = \sin A \cos B + \cos A \sin B \), we get:
\( \sin 2x = \sin((x+\frac{\pi}{6}) + (x-\frac{\pi}{6})) = \sin(x+\frac{\pi}{6})\cos(x-\frac{\pi}{6}) + \cos(x+\frac{\pi}{6})\sin(x-\frac{\pi}{6}) \).
So, the integral becomes:
\( I = \int \frac{\sin(x+\frac{\pi}{6})\cos(x-\frac{\pi}{6}) + \cos(x+\frac{\pi}{6})\sin(x-\frac{\pi}{6})}{\sin(x-\frac{\pi}{6})\sin(x+\frac{\pi}{6})} dx \)
Now, we split this into two fractions:
\( = \int \left( \frac{\cos(x-\frac{\pi}{6})}{\sin(x-\frac{\pi}{6})} + \frac{\cos(x+\frac{\pi}{6})}{\sin(x+\frac{\pi}{6})} \right) dx \)
\( = \int \left( \cot(x-\frac{\pi}{6}) + \cot(x+\frac{\pi}{6}) \right) dx \)
We integrate each term separately:
\( = \log |\sin(x-\frac{\pi}{6})| + \log |\sin(x+\frac{\pi}{6})| + C \)
Using the logarithm property \( \log A + \log B = \log (AB) \):
\( = \log |\sin(x-\frac{\pi}{6})\sin(x+\frac{\pi}{6})| + C \)
In simple words: For both parts, we changed the top part of the fraction using trigonometric identities. This helped us split the integral into simpler parts that use the cotangent function. Integrating cotangent gives a logarithm involving sine, which is a common pattern in integration.

🎯 Exam Tip: When the numerator is a sine of a multiple angle and the denominator is a product of sines, try to express the numerator as the sum or difference of the denominator angles to simplify the fraction into cotangent terms.

Question 14. Find the following:
Answer: We need to integrate \( I = \int \frac{1}{3\sin x + 4\cos x} dx \).
To solve this, we use a trick: express the denominator in the form \( r \sin(x+\theta) \).
Let \( 3 = r \cos\theta \) and \( 4 = r \sin\theta \).
Then \( r^2 = (r \cos\theta)^2 + (r \sin\theta)^2 = 3^2 + 4^2 = 9 + 16 = 25 \). So, \( r = \sqrt{25} = 5 \).
Also, \( \frac{r \sin\theta}{r \cos\theta} = \frac{4}{3} \implies \tan\theta = \frac{4}{3} \). So, \( \theta = \tan^{-1}\left(\frac{4}{3}\right) \).
The denominator becomes \( 3\sin x + 4\cos x = r \cos\theta \sin x + r \sin\theta \cos x = r(\sin x \cos\theta + \cos x \sin\theta) = r \sin(x+\theta) \).
Substitute this back into the integral:
\( I = \int \frac{1}{r \sin(x+\theta)} dx = \frac{1}{r} \int \frac{1}{\sin(x+\theta)} dx \)
We know that \( \frac{1}{\sin A} = \operatorname{cosec} A \).
\( I = \frac{1}{r} \int \operatorname{cosec}(x+\theta) dx \)
The integral of \( \operatorname{cosec} A \) is \( \log |\operatorname{cosec} A - \cot A| \).
\( = \frac{1}{r} \log |\operatorname{cosec}(x+\theta) - \cot(x+\theta)| + C \)
Substitute \( r = 5 \) back:
\( = \frac{1}{5} \log |\operatorname{cosec}(x+\theta) - \cot(x+\theta)| + C \)
Alternatively, using the formula \( \operatorname{cosec} A - \cot A = \tan \frac{A}{2} \):
\( = \frac{1}{5} \log \left| \tan \left( \frac{x+\theta}{2} \right) \right| + C \)
Finally, substitute \( \theta = \tan^{-1}\left(\frac{4}{3}\right) \):
\( = \frac{1}{5} \log \left| \tan \left( \frac{x+\tan^{-1}\left(\frac{4}{3}\right)}{2} \right) \right| + C \)
In simple words: To solve this integral, we first rewrote the bottom part of the fraction as a single sine function using a clever trigonometric substitution. This made the integral much simpler, turning it into the integral of cosecant. Then we used the standard formula for integrating cosecant, and put back the values.

🎯 Exam Tip: For integrals of the form \( \frac{1}{a \sin x + b \cos x} \), always convert the denominator to \( r \sin(x+\theta) \) or \( r \cos(x-\theta) \) using \( a = r \cos\theta \) and \( b = r \sin\theta \). This transformation is key to simplifying the integral.

Question 15. Find the following:
(a) \( \int \frac{\sin x \cos x}{a \cos^2 x + b \sin^2 x} dx \)
(b) \( \int \frac{\sec x}{\sqrt{\sin (2x+a) + \sin a}} dx \)
Answer:
(a) We need to integrate \( I = \int \frac{\sin x \cos x}{a \cos^2 x + b \sin^2 x} dx \).
We use the substitution method. Let the denominator be \( t \).
Let \( t = a \cos^2 x + b \sin^2 x \).
Now, we find the derivative of \( t \) with respect to \( x \):
\( \frac{dt}{dx} = a \cdot (2 \cos x (-\sin x)) + b \cdot (2 \sin x \cos x) \)
\( \frac{dt}{dx} = -2a \sin x \cos x + 2b \sin x \cos x \)
\( \frac{dt}{dx} = 2(b-a) \sin x \cos x \)
So, \( dt = 2(b-a) \sin x \cos x dx \).
This means \( \sin x \cos x dx = \frac{1}{2(b-a)} dt \).
Substitute \( t \) and \( dt \) back into the integral:
\( I = \int \frac{1}{t} \left( \frac{1}{2(b-a)} dt \right) \)
\( = \frac{1}{2(b-a)} \int \frac{1}{t} dt \)
The integral of \( \frac{1}{t} \) is \( \log |t| \).
\( = \frac{1}{2(b-a)} \log |t| + C \)
Finally, substitute \( t = a \cos^2 x + b \sin^2 x \) back:
\( = \frac{1}{2(b-a)} \log |a \cos^2 x + b \sin^2 x| + C \)

(b) We need to integrate \( I = \int \frac{\sec x}{\sqrt{\sin (2x+a) + \sin a}} dx \).
First, simplify the denominator using the sum-to-product identity: \( \sin C + \sin D = 2 \sin\left(\frac{C+D}{2}\right)\cos\left(\frac{C-D}{2}\right) \).
Here, \( C = 2x+a \) and \( D = a \).
\( \frac{C+D}{2} = \frac{2x+a+a}{2} = x+a \)
\( \frac{C-D}{2} = \frac{2x+a-a}{2} = x \)
So, \( \sin (2x+a) + \sin a = 2 \sin(x+a)\cos x \).
The integral becomes:
\( I = \int \frac{\sec x}{\sqrt{2 \sin(x+a)\cos x}} dx \)
Replace \( \sec x \) with \( \frac{1}{\cos x} \):
\( = \int \frac{1/\cos x}{\sqrt{2 \sin(x+a)\cos x}} dx \)
Multiply the numerator and denominator by \( \cos x \) to bring \( \sec^2 x \) to the top and \( \cos^2 x \) under the root:
\( = \int \frac{\sec^2 x \cdot \sqrt{\cos x}}{\sqrt{2 \sin(x+a)\cos x} \cdot \sqrt{\cos x}} dx \)
This leads to:
\( = \frac{1}{\sqrt{2}} \int \frac{\sec^2 x}{\sqrt{\sin(x+a)}} dx \)
Expand \( \sin(x+a) = \sin x \cos a + \cos x \sin a \):
\( = \frac{1}{\sqrt{2}} \int \frac{\sec^2 x}{\sqrt{\sin x \cos a + \cos x \sin a}} dx \)
Divide the numerator and denominator of the fraction inside the integral by \( \cos x \) (to change the terms under the square root into tangent form):
\( = \frac{1}{\sqrt{2}} \int \frac{\sec^2 x / \sqrt{\cos^2 x}}{\sqrt{(\sin x \cos a + \cos x \sin a) / \cos x}} dx \)
\( = \frac{1}{\sqrt{2}} \int \frac{\sec^2 x}{\sqrt{\tan x \cos a + \sin a}} dx \)
Now, we use substitution. Let \( t = \tan x \cos a + \sin a \).
Differentiate \( t \) with respect to \( x \):
\( \frac{dt}{dx} = (\sec^2 x \cos a) + 0 \)
So, \( dt = \sec^2 x \cos a dx \), which means \( \sec^2 x dx = \frac{dt}{\cos a} \).
Substitute \( t \) and \( dt \) into the integral:
\( I = \frac{1}{\sqrt{2}} \int \frac{1}{\sqrt{t}} \left( \frac{dt}{\cos a} \right) \)
\( = \frac{1}{\sqrt{2}\cos a} \int t^{-1/2} dt \)
Integrate \( t^{-1/2} \):
\( = \frac{1}{\sqrt{2}\cos a} \left( \frac{t^{-1/2+1}}{-1/2+1} \right) + C \)
\( = \frac{1}{\sqrt{2}\cos a} \left( \frac{t^{1/2}}{1/2} \right) + C \)
\( = \frac{2}{\sqrt{2}\cos a} \sqrt{t} + C \)
Simplify \( \frac{2}{\sqrt{2}} \) to \( \sqrt{2} \):
\( = \frac{\sqrt{2}}{\cos a} \sqrt{t} + C \)
Substitute \( t = \tan x \cos a + \sin a \) back and note \( \frac{1}{\cos a} = \sec a \):
\( = \sqrt{2} \sec a \sqrt{\tan x \cos a + \sin a} + C \)
In simple words: For part (a), we noticed the top part of the fraction was almost the derivative of the bottom part, so we used a simple substitution. For part (b), we used a trigonometric identity to simplify the expression under the square root, and then another substitution helped us solve it.

🎯 Exam Tip: Look for opportunities to use substitution when one part of the integral is the derivative of another. For trigonometric integrals, identities like sum-to-product can simplify complex denominators or numerators significantly.

Question 16. Find the following:
(a) \( \int \frac{1}{\cos^3 x \sin(x+a)} dx \)
(b) \( \int \frac{\cos 2x - \cos 2a}{\cos x - \cos a} dx \)
Answer:
(a) This answer provides the solution for the integral \( \int \frac{\sec^2 x}{\sqrt{\tan x \cos a + \sin a}} dx \).
To solve this integral, we use substitution. Let \( t = \tan x \cos a + \sin a \).
Now, we find the derivative of \( t \) with respect to \( x \):
\( \frac{dt}{dx} = \sec^2 x \cos a \).
So, \( dt = \sec^2 x \cos a dx \). This means \( \sec^2 x dx = \frac{dt}{\cos a} \).
Substitute \( t \) and \( dt \) into the integral:
\( = \int \frac{1}{\sqrt{t}} \left( \frac{dt}{\cos a} \right) \)
\( = \frac{1}{\cos a} \int t^{-1/2} dt \)
Integrate \( t^{-1/2} \):
\( = \frac{1}{\cos a} \left( \frac{t^{-1/2+1}}{-1/2+1} \right) + C \)
\( = \frac{1}{\cos a} \left( \frac{t^{1/2}}{1/2} \right) + C \)
\( = \frac{2}{\cos a} \sqrt{t} + C \)
Finally, substitute \( t = \tan x \cos a + \sin a \) back:
\( = \frac{2}{\cos a} \sqrt{\tan x \cos a + \sin a} + C \)

(b) We need to integrate \( I = \int \frac{\cos 2x - \cos 2a}{\cos x - \cos a} dx \).
First, we simplify the numerator using the double-angle identity \( \cos 2A = 2\cos^2 A - 1 \).
So, \( \cos 2x - \cos 2a = (2\cos^2 x - 1) - (2\cos^2 a - 1) \).
\( = 2\cos^2 x - 1 - 2\cos^2 a + 1 \)
\( = 2\cos^2 x - 2\cos^2 a \)
\( = 2(\cos^2 x - \cos^2 a) \)
Now, use the difference of squares identity \( A^2 - B^2 = (A-B)(A+B) \):
\( = 2(\cos x - \cos a)(\cos x + \cos a) \).
Substitute this back into the integral:
\( I = \int \frac{2(\cos x - \cos a)(\cos x + \cos a)}{\cos x - \cos a} dx \)
Assuming \( \cos x - \cos a \neq 0 \), we can cancel the term:
\( = \int 2(\cos x + \cos a) dx \)
Split the integral:
\( = 2 \int \cos x dx + 2 \int \cos a dx \)
Integrate each term. \( \cos a \) is a constant with respect to \( x \):
\( = 2 \sin x + 2x \cos a + C \)
We can factor out 2:
\( = 2(\sin x + x \cos a) + C \)
In simple words: For part (a), the solution provided solves a different integral than stated. It uses a simple substitution to integrate a function involving tangent and secant. For part (b), we used trigonometric identities to simplify the top part of the fraction, allowing us to cancel a term with the bottom part. This made the integral much easier to solve directly.

🎯 Exam Tip: When dealing with integrals involving \( \cos 2x \) and \( \cos 2a \), remember to use the double-angle identity \( \cos 2\theta = 2\cos^2 \theta - 1 \) to simplify the expression, often leading to a difference of squares that can be factored.