₀²π} m x n x dx + _{-π}^{π} m x n x dx =

Correct Answer is : πππ if m=n,m,n∈Zm=n, m, n Zm=n,m,n∈Z

Correct Answer is : πππ if m=n,m,n∈Zm=n, m, n Zm=n,m,n∈Z

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AP EAPCET 08-Jul-2022 Shift 1 Paper

Section: Mathematics

Question : 75 of 160

Marks: +1, -0

\int\limits_{0}^{2\pi} \cos m x \cos n x dx + \int\limits_{-\pi}^{\pi} \sin m x \cos n x dx =

Solution: 👈: Video Solution

When,

m=n, m \in Z

Let

I = \int\limits_{0}^{2\pi} \cos^{2}(m x) \, dx + \frac{1}{2} \int\limits_{-\pi}^{\pi} 2 \sin(m x) \cos(m x) \, dx

\int\limits_{0}^{2\pi} \cos^2 m x dx + \frac{1}{2} \int\limits_{-\pi}^{\pi} \sin 2 m x dx

Using the formula.

\int\limits_{0}^{2a} f(x) = \begin{cases} 2\int\limits_{0}^{a} f(x) dx & \text{if } f(2a-x)=f(x) \\ 0 & \text{if } f(2a-x)=-f(x) \end{cases}

and

\int\limits_{-a}^{a} f(x) = \begin{cases} 2\int\limits_{0}^{a} f(x) dx & \text{if } f(-x)=f(x) \\ 0 & \text{if } f(-x)=-f(x) \end{cases}

I = 2 \int\limits_{0}^{\pi} \cos^2 m x dx + 0

= \int_{0}^{\pi} (1 + \cos 2 m x) \, dx

= \left[ x + \frac{\sin 2 m x}{2 m} \right]_0^{\pi} = \pi

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