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Exercise 11.3 · Q2

Q.Integrate the following with respect to xx:
[!FORMULA] 4cos⁡(5−2x)+9e3x−6+246−4x4\cos(5-2x)+9e^{3x-6}+\dfrac{24}{6-4x}

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✓ Free question

Split the sum into three standard forms — cosine, exponential, and log⁡\log — each on a linear argument, apply §11.4 to each, and add.

Step 1. Cosine term. a=−2a=-2; ∫cos⁡x dx=sin⁡x+c\int\cos x\,dx=\sin x+c, so

∫4cos⁡(5−2x) dx=4×1−2sin⁡(5−2x)+c=−2sin⁡(5−2x)+c.\int 4\cos(5-2x)\,dx=4\times\frac{1}{-2}\sin(5-2x)+c=-2\sin(5-2x)+c.

Step 2. Exponential term. a=3a=3; so

∫9e3x−6 dx=9×13e3x−6+c=3e3x−6+c.\int 9e^{3x-6}\,dx = 9\times\frac13 e^{3x-6}+c=3e^{3x-6}+c.

Step 3. Reciprocal-linear term. a=−4a=-4; so

∫246−4x dx=24×1−4log⁡∣6−4x∣+c=−6log⁡∣6−4x∣+c.\int \frac{24}{6-4x}\,dx = 24\times\frac{1}{-4}\log|6-4x|+c=-6\log|6-4x|+c.

Step 4. Combine.

−2sin⁡(5−2x)+3e3x−6−6log⁡∣6−4x∣+c.-2\sin(5-2x)+3e^{3x-6}-6\log|6-4x|+c.

Step 5. Check by differentiating. ddx(−2sin⁡(5−2x))=−2cos⁡(5−2x)⋅(−2)=4cos⁡(5−2x)\dfrac{d}{dx}\big(-2\sin(5-2x)\big)=-2\cos(5-2x)\cdot(-2)=4\cos(5-2x) ✓. ddx(3e3x−6)=9e3x−6\dfrac{d}{dx}\big(3e^{3x-6}\big)=9e^{3x-6} ✓. ddx(−6log⁡∣6−4x∣)=−6⋅−46−4x=246−4x\dfrac{d}{dx}\big(-6\log|6-4x|\big)=-6\cdot\dfrac{-4}{6-4x}=\dfrac{24}{6-4x} ✓.

✓Final answer

−2sin⁡(5−2x)+3e3x−6−6log⁡∣6−4x∣+c-2\sin(5-2x)+3e^{3x-6}-6\log|6-4x|+c

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