Solveeit Logo
Login

Question

Mathematics Question on Integration

The integral ∫1/43/4cos⁡(2cot⁡−11−x1+x) dx\int_{1/4}^{3/4} \cos\left( 2 \cot^{-1} \sqrt{\frac{1 - x}{1 + x}} \right) \, dx is equal to:

A

−12-\frac{1}{2}

B

14\frac{1}{4}

C

12\frac{1}{2}

D

−14-\frac{1}{4}

Answer

−14-\frac{1}{4}

Explanation

Solution

Given:

θ=cot⁡−1(1−x1+x)\theta = \cot^{-1} \left( \sqrt{\frac{1 - x}{1 + x}} \right)

Therefore:

cot⁡(θ)=1−x1+x  ⟹  tan⁡(θ)=1+x1−x\cot(\theta) = \sqrt{\frac{1 - x}{1 + x}} \implies \tan(\theta) = \sqrt{\frac{1 + x}{1 - x}}

Using the double-angle formula for cosine:

cos⁡(2θ)=1−2sin⁡2(θ)\cos(2\theta) = 1 - 2\sin^2(\theta)

Now express sin⁡2(θ)\sin^2(\theta):

sin⁡2(θ)=11+cot⁡2(θ)\sin^2(\theta) = \frac{1}{1 + \cot^2(\theta)}

Substituting cot⁡2(θ)=1−x1+x\cot^2(\theta) = \frac{1 - x}{1 + x}, we get:

sin⁡2(θ)=11+1−x1+x=1+x2\sin^2(\theta) = \frac{1}{1 + \frac{1 - x}{1 + x}} = \frac{1 + x}{2}

Thus:

cos⁡(2θ)=1−2sin⁡2(θ)=1−2⋅1+x2=−x\cos(2\theta) = 1 - 2\sin^2(\theta) = 1 - 2 \cdot \frac{1 + x}{2} = -x

The integral simplifies to:

∫1/43/4cos⁡(2cot⁡−11−x1+x)dx=∫1/43/4−xdx\int_{1/4}^{3/4} \cos \left( 2 \cot^{-1} \sqrt{\frac{1 - x}{1 + x}} \right) dx = \int_{1/4}^{3/4} -x dx

Evaluate the simplified integral:

∫1/43/4−xdx=−∫1/43/4xdx\int_{1/4}^{3/4} -x dx = - \int_{1/4}^{3/4} x dx

The integral of xx is:

∫xdx=x22\int x dx = \frac{x^2}{2}

Evaluate the limits:

\-[x22]1/43/4=−((34)22−(14)22)\- \left[ \frac{x^2}{2} \right]_{1/4}^{3/4} = - \left( \frac{\left(\frac{3}{4}\right)^2}{2} - \frac{\left(\frac{1}{4}\right)^2}{2} \right)

Simplify:

\-(932−132)=−832=−14\- \left( \frac{9}{32} - \frac{1}{32} \right) = - \frac{8}{32} = -\frac{1}{4}

Final Answer:

−14-\frac{1}{4}