Math, asked by PragyaTbia, 1 year ago

Evaluate the definite integrals: \int^\frac{\pi}{4}_0 {sin\ 2x} \, dx

Answers

Answered by MaheswariS
0

Answer:

\frac{1}{2}

Step-by-step explanation:

Concept:

\int{sinax}\:dx=\frac{-1}{a}cosax+c

Now,

\int\limits^{\frac{\pi}{4}}_0{sin2x}\:dx \\\\=[\frac{-cos2x}{2}]^{\frac{\pi}{4}}_0 \\\\=\frac{-1}{2}[cos2x]^{\frac{\pi}{4}}_0 \\\\=\frac{-1}{2}[cos2(\frac{\pi}{4})-cos2(0)] \\\\=\frac{-1}{2}[cos(\frac{\pi}{2})-cos(0)] \\\\=\frac{-1}{2}[0-1] \\\\=\frac{1}{2}

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