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Exercise 11.2 · Q2

Q.Integrate the following functions with respect to xx:

(i) sin⁡3x\sin 3x
(ii) cos⁡(5−11x)\cos(5-11x)
(iii) cosec2(5x−7)\text{cosec}^{2}(5x-7)
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Each part is a standard trig form (sin⁡\sin, cos⁡\cos, cosec2\text{cosec}^2) evaluated at a linear argument ax+bax+b; integrate as usual and divide by aa.

Step 1. Part (i). a=3a=3; ∫sin⁡x dx=−cos⁡x+c\int\sin x\,dx=-\cos x+c, so

∫sin⁡3x dx=−13cos⁡3x+c.\int \sin 3x\,dx = -\frac13\cos 3x+c.

Step 2. Part (ii). 5−11x5-11x has a=−11a=-11; ∫cos⁡x dx=sin⁡x+c\int\cos x\,dx=\sin x+c, so

∫cos⁡(5−11x) dx=1−11sin⁡(5−11x)+c=−111sin⁡(5−11x)+c.\int \cos(5-11x)\,dx = \frac{1}{-11}\sin(5-11x)+c=-\frac{1}{11}\sin(5-11x)+c.

Step 3. Part (iii). a=5a=5; ∫cosec2x dx=−cot⁡x+c\int\text{cosec}^2x\,dx=-\cot x+c, so

∫cosec2(5x−7) dx=15(−cot⁡(5x−7))+c=−15cot⁡(5x−7)+c.\int \text{cosec}^2(5x-7)\,dx = \frac15\big(-\cot(5x-7)\big)+c=-\frac15\cot(5x-7)+c.

Step 4. Check by differentiating. ddx ⁣(−13cos⁡3x)=13sin⁡3x⋅3=sin⁡3x\dfrac{d}{dx}\!\left(-\dfrac13\cos3x\right)=\dfrac13\sin3x\cdot3=\sin3x ✓. ddx ⁣(−111sin⁡(5−11x))=−111cos⁡(5−11x)⋅(−11)=cos⁡(5−11x)\dfrac{d}{dx}\!\left(-\dfrac{1}{11}\sin(5-11x)\right)=-\dfrac{1}{11}\cos(5-11x)\cdot(-11)=\cos(5-11x) ✓. ddx ⁣(−15cot⁡(5x−7))=15cosec2(5x−7)⋅5=cosec2(5x−7)\dfrac{d}{dx}\!\left(-\dfrac15\cot(5x-7)\right)=\dfrac15\text{cosec}^2(5x-7)\cdot5=\text{cosec}^2(5x-7) ✓.

✓Final answer

(i) −13cos⁡3x+c-\dfrac13\cos 3x+c (ii) −111sin⁡(5−11x)+c-\dfrac{1}{11}\sin(5-11x)+c (iii) −15cot⁡(5x−7)+c-\dfrac15\cot(5x-7)+c

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