Class 12 Maths, Worked in Full Past papersGet updates
Chapter 7 · Integrals

Miscellaneous · Q15: ∫cos⁡3x elog⁡sin⁡x dx\int\cos^3x\,e^{\log\sin x}\,dx

Miscellaneous ExerciseSimplify e^(log u) = u, then substitute t = cos x · Miscellaneous Exercise
The question

Integrate the function:

cos⁡3x  elog⁡sin⁡x\cos^3x\; e^{\log\sin x}
Exam answer

Let I=∫cos⁡3x elog⁡sin⁡x dxI = \displaystyle\int\cos^3 x\,e^{\log\sin x}\,dx⁠.

Since sin⁡x>0\sin x > 0⁠, elog⁡sin⁡x=sin⁡xe^{\log\sin x} = \sin x⁠. So

I=∫cos⁡3xsin⁡x dxI = \int\cos^3 x\sin x\,dx

Put cos⁡x=t\cos x = t⁠. Then −sin⁡x dx=dt-\sin x\,dx = dt⁠.

I=−∫t3 dt=−t44+CI = -\int t^3\,dt = -\frac{t^4}{4} + C

∴ I=−14cos⁡4x+C\therefore\ I = -\dfrac14\cos^4 x + C⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The exponential and the logarithm undo each other: elog⁡u=ue^{\log u} = u for every u>0u > 0⁠. So the frightening factor elog⁡sin⁡xe^{\log\sin x} is just sin⁡x\sin x (wherever it is defined), and the integrand is cos⁡3xsin⁡x\cos^3x\sin x⁠. Since sin⁡x dx\sin x\,dx is minus the differential of cos⁡x\cos x⁠, the substitution t=cos⁡xt = \cos x finishes it.

Given, to find, and the plan
Given
f(x)=cos⁡3x elog⁡sin⁡xf(x) = \cos^3x\,e^{\log\sin x}⁠. The logarithm needs sin⁡x>0\sin x > 0⁠, so ff is defined exactly on the intervals (2kπ,(2k+1)π)(2k\pi, (2k+1)\pi)⁠, kk an integer.
To find
∫cos⁡3x elog⁡sin⁡x dx\displaystyle\int\cos^3x\,e^{\log\sin x}\,dx⁠.
Plan
(1) Simplify elog⁡sin⁡xe^{\log\sin x}⁠. (2) Choose the substitution t=cos⁡xt = \cos x⁠. (3) Integrate the power of tt⁠. (4) Return to xx⁠. (5) Differentiate back and compare with the key.

Basics you need

1Exponential and logarithm are inverses

elog⁡u=ue^{\log u} = u for u>0u > 0⁠; log⁡u\log u does not exist for u≤0u \le 0⁠. Basics: logarithms and 1/x

2Substitution (Section 7.3.1)

If t=g(x)t = g(x)⁠, then dt=g′(x) dxdt = g'(x)\,dx⁠, and ∫h(g(x)) g′(x) dx=∫h(t) dt\int h(g(x))\,g'(x)\,dx = \int h(t)\,dt⁠. Basics: substitution

3Power formula

∫tn dt=tn+1n+1+C\int t^n\,dt = \frac{t^{n+1}}{n+1} + C for n≠−1n \ne -1 (standard formula (i), Section 7.2). Basics: powers of x

Solution

  1. Simplify elog⁡sin⁡xe^{\log\sin x}

    On the domain, sin⁡x>0\sin x > 0⁠, so

    elog⁡sin⁡x=sin⁡x,f(x)=cos⁡3xsin⁡x.e^{\log\sin x} = \sin x ,\qquad f(x) = \cos^3x\sin x .
    log⁡\log is the inverse of the exponential function: log⁡u\log u is the power to which ee must be raised to give uu⁠. Raising ee to that power gives uu back. This only makes sense when log⁡sin⁡x\log\sin x exists, that is when sin⁡x>0\sin x > 0⁠; that is why the domain is the set of intervals where sin⁡x\sin x is positive, not all of R\mathbb{R}⁠.
    Basics: logarithms and 1/x

  2. Choose the method: substitute t=cos⁡xt = \cos x

    t=cos⁡x,dt=−sin⁡x dx,sin⁡x dx=−dt.t = \cos x ,\qquad dt = -\sin x\,dx ,\qquad \sin x\,dx = -dt .
    The integrand is a power of cos⁡x\cos x multiplied by sin⁡x\sin x⁠, and sin⁡x\sin x is (up to sign) the derivative of cos⁡x\cos x⁠. That is exactly the pattern h(g(x)) g′(x)h(g(x))\,g'(x) that substitution handles, with g(x)=cos⁡xg(x) = \cos x and h(t)=−t3h(t) = -t^3⁠.
    Section 7.3.1. Lesson: integration by substitution

  3. Integrate in tt

    ∫cos⁡3xsin⁡x dx=∫t3 (−dt)=−t44+C.\int\cos^3x\sin x\,dx = \int t^3\,(-dt) = -\frac{t^4}{4} + C .
    Power formula with n=3n = 3⁠; the minus sign comes from sin⁡x dx=−dt\sin x\,dx = -dt⁠.
    Standard formula (i), Section 7.2. Basics: powers of x

  4. Return to xx

    ∫cos⁡3x elog⁡sin⁡x dx=−14cos⁡4x+C.\int\cos^3x\,e^{\log\sin x}\,dx = -\frac14\cos^4x + C .
    Replace tt by cos⁡x\cos x⁠. The answer must be written in the variable we started with.

  5. Check: differentiate the answer

    ddx(−14cos⁡4x)=−14⋅4cos⁡3x⋅(−sin⁡x)=cos⁡3xsin⁡x=cos⁡3x elog⁡sin⁡x ✓\begin{aligned}\frac{d}{dx}\left(-\frac14\cos^4x\right) &= -\frac14\cdot 4\cos^3x\cdot(-\sin x)\\ &= \cos^3x\sin x\\ &= \cos^3x\,e^{\log\sin x}\ \checkmark\end{aligned}
    Chain rule: outer function u4u^4⁠, inner function u=cos⁡xu = \cos x with derivative −sin⁡x-\sin x⁠. The last equality is Step 1, valid where sin⁡x>0\sin x > 0⁠. So the answer is valid on each interval (2kπ,(2k+1)π)(2k\pi, (2k+1)\pi)⁠.
    Basics: integration undoes differentiation

  6. Compare with the textbook, and a look-alike answer

    The key gives −14cos⁡4x+C-\frac14\cos^4x + C⁠, exactly our answer.

    A student who substitutes t=sin⁡xt = \sin x instead (writing cos⁡2x=1−sin⁡2x\cos^2x = 1 - \sin^2x⁠) gets 12sin⁡2x−14sin⁡4x+C\frac12\sin^2x - \frac14\sin^4x + C⁠. This is also correct:

    −14cos⁡4x=−14(1−sin⁡2x)2=−14+12sin⁡2x−14sin⁡4x,\begin{aligned}-\frac14\cos^4x &= -\frac14\left(1 - \sin^2x\right)^2\\ &= -\frac14 + \frac12\sin^2x - \frac14\sin^4x ,\end{aligned}

    so the two answers differ only by the constant −14-\frac14⁠.

    Two antiderivatives of the same function on an interval differ by a constant (Remark in Section 7.2), and the arbitrary CC absorbs it.
    Remark, Section 7.2. Basics: trigonometric identities
Result

Where sin⁡x>0\sin x > 0⁠:

∫cos⁡3x elog⁡sin⁡x dx=−14cos⁡4x+C\int\cos^3x\,e^{\log\sin x}\,dx = -\frac14\cos^4x + C

(Textbook answer: −14cos⁡4x+C-\frac14\cos^4x + C⁠.)

Check it with numbers

On [π6,π2][\frac{\pi}{6}, \frac{\pi}{2}] we have sin⁡x≥12>0\sin x \ge \frac12 > 0⁠, so the integrand is continuous there. With FF our answer,

F(π2)−F(π6)=0+14(32)4=964=0.140625.F\left(\tfrac{\pi}{2}\right) - F\left(\tfrac{\pi}{6}\right) = 0 + \frac14\left(\frac{\sqrt3}{2}\right)^4 = \frac{9}{64} = 0.140625 .

Numerical integration of cos⁡3x elog⁡sin⁡x\cos^3x\,e^{\log\sin x} from π6\frac{\pi}{6} to π2\frac{\pi}{2} also gives 0.1406250.140625⁠.

Common mistakes
  1. Writing elog⁡sin⁡x=log⁡sin⁡xe^{\log\sin x} = \log\sin x or esin⁡xe^{\sin x}⁠. The exponential undoes the logarithm and leaves sin⁡x\sin x⁠.
  2. Losing the minus sign. dt=−sin⁡x dxdt = -\sin x\,dx⁠, so the answer is −14cos⁡4x-\frac14\cos^4x⁠, not +14cos⁡4x+\frac14\cos^4x⁠.
  3. Claiming the answer holds for all xx⁠. The given function exists only where sin⁡x>0\sin x > 0⁠.
What to take away
  1. Simplify first: elog⁡u=ue^{\log u} = u (for u>0u > 0⁠), and similarly eklog⁡u=uke^{k\log u} = u^k⁠.
  2. ∫cos⁡nxsin⁡x dx=−cos⁡n+1xn+1+C\int\cos^n x\sin x\,dx = -\frac{\cos^{n+1}x}{n+1} + C for n≠−1n \ne -1⁠, by t=cos⁡xt = \cos x⁠.
  3. Different-looking answers can both be right; their difference must be a constant.

Every fact used, and where it comes from

FactUsed inSource
elog⁡u=ue^{\log u} = u for u>0u > 0Step 1Chapter 5, exponential and logarithmic functions (Basics)
Integration by substitutionSteps 2, 3Section 7.3.1 (Lesson: substitution)
∫tn dt=tn+1n+1+C\int t^n\,dt = \frac{t^{n+1}}{n+1} + CStep 3Standard formula (i), Section 7.2
Chain rule; ddxcos⁡x=−sin⁡x\frac{d}{dx}\cos x = -\sin xStep 5Chapter 5
Antiderivatives differ by a constantStep 6Remark, Section 7.2
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →