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Q.Find ∫ecot⁡−1x(1−x+x21+x2)dx\int e^{\cot^{-1}x}\left(\frac{1 - x + x^2}{1 + x^2}\right)dx.

CBSECBSE Class XII Board 2023Subjective· 3mImportance★★★★★
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By substituting t=cot⁡−1xt = \cot^{-1}x, the integral transforms into the standard form ∫et(f(t)+f′(t))dt\int e^t (f(t) + f'(t)) dt, which evaluates to etf(t)+Ce^t f(t) + C, leading to the final result xecot⁡−1x+Cx e^{\cot^{-1}x} + C.

Concept and Intuition

When an integral contains an exponential term with a complex exponent, such as ecot⁡−1xe^{\cot^{-1}x}, a natural first step is often to simplify this exponent through a substitution. Let t=cot⁡−1xt = \cot^{-1}x. This substitution aims to make the exponential part simpler (ete^t) and hopefully transform the rest of the integrand into a more recognizable form.

A key observation here is the presence of 11+x2\frac{1}{1+x^2} in the denominator of the given fraction. This term is directly related to the derivative of cot⁡−1x\cot^{-1}x, which is −11+x2\frac{-1}{1+x^2}. This connection strongly suggests that the substitution t=cot⁡−1xt = \cot^{-1}x will be effective, as it will simplify the dxdx term.

After performing the substitution and simplifying the algebraic expression in terms of tt, our goal is to see if the integral matches a known pattern. A very powerful and frequently tested pattern is ∫eu(f(u)+f′(u))du=euf(u)+C\int e^u (f(u) + f'(u)) du = e^u f(u) + C. If we can manipulate the integrand into this specific form, the solution becomes straightforward.

Let's proceed with the substitution and the necessary algebraic and trigonometric simplifications.

  1. Perform the substitution:

    Let t=cot⁡−1xt = \cot^{-1}x.

    From this, we can express xx in terms of tt: x=cot⁡tx = \cot t.

    Next, we need to find dxdx in terms of dtdt. Differentiate t=cot⁡−1xt = \cot^{-1}x with respect to xx:

dtdx=−11+x2\frac{dt}{dx} = \frac{-1}{1+x^2}

Rearranging this gives us the expression for $dx$:

dx=−(1+x2)dtdx = -(1+x^2) dt

  1. Transform the integrand into terms of tt: Substitute t=cot⁡−1xt = \cot^{-1}x, x=cot⁡tx = \cot t, and dx=−(1+x2)dtdx = -(1+x^2) dt into the original integral:

I=∫ecot⁡−1x(1−x+x21+x2)dxI = \int e^{\cot^{-1}x}\left(\frac{1 - x + x^2}{1 + x^2}\right)dx

I=∫et(1−cot⁡t+cot⁡2t1+cot⁡2t)(−(1+x2))dtI = \int e^t \left(\frac{1 - \cot t + \cot^2 t}{1 + \cot^2 t}\right) (-(1+x^2)) dt

We know that $1+x^2 = 1+\cot^2 t$. Using the trigonometric identity $1+\cot^2 t = \csc^2 t$, we can simplify this term.
So, $-(1+x^2)$ becomes $-\csc^2 t$.

Now, let's simplify the fraction $\frac{1 - \cot t + \cot^2 t}{1 + \cot^2 t}$:
We can rewrite the numerator as $(1 + \cot^2 t) - \cot t$.

1−cot⁡t+cot⁡2t1+cot⁡2t=(1+cot⁡2t)−cot⁡t1+cot⁡2t\frac{1 - \cot t + \cot^2 t}{1 + \cot^2 t} = \frac{(1 + \cot^2 t) - \cot t}{1 + \cot^2 t}

Substitute $1+\cot^2 t = \csc^2 t$:

=csc⁡2t−cot⁡tcsc⁡2t= \frac{\csc^2 t - \cot t}{\csc^2 t}

Separate the terms:

=csc⁡2tcsc⁡2t−cot⁡tcsc⁡2t=1−cot⁡tcsc⁡2t= \frac{\csc^2 t}{\csc^2 t} - \frac{\cot t}{\csc^2 t} = 1 - \frac{\cot t}{\csc^2 t}

Now, simplify $\frac{\cot t}{\csc^2 t}$:

cot⁡tcsc⁡2t=cos⁡t/sin⁡t1/sin⁡2t=cos⁡tsin⁡t⋅sin⁡2t=sin⁡tcos⁡t\frac{\cot t}{\csc^2 t} = \frac{\cos t / \sin t}{1 / \sin^2 t} = \frac{\cos t}{\sin t} \cdot \sin^2 t = \sin t \cos t

So, the fraction simplifies to $1 - \sin t \cos t$.

Substitute these simplified terms back into the integral expression:

I=∫et(1−sin⁡tcos⁡t)(−csc⁡2t)dtI = \int e^t (1 - \sin t \cos t) (-\csc^2 t) dt

Distribute $-\csc^2 t$:

I=∫et(−csc⁡2t+sin⁡tcos⁡tcsc⁡2t)dtI = \int e^t (-\csc^2 t + \sin t \cos t \csc^2 t) dt

Recall that $\csc^2 t = \frac{1}{\sin^2 t}$:
$$I = \int e^t \left(-\csc^2 t + \sin t \cos t \frac{1}{\sin^2 t}\right) dt$$ …
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