Question

Evaluate the integral
integration [(sin2x
+ Cos2x)dx] between the limits x= PI/4
and x= PI/3. Answer should be correct to one decimal​

Answer 1

$\large\underline{\bold{Solution-}}$

$\rm :\longmapsto\:Let \: I \: = \: \int_ \frac{\pi}{4} ^ \frac{\pi}{3} \: (sin2x \: + \: cos2x) \: dx$

We know that

$\boxed{ \sf{ \int \: sinx \: dx \: = \: - \: cosx \: + \: c}}$

and

$\boxed{ \sf{ \int \: cosx \: dx \: = \: sinx \: + \: c}}$

So,

$\rm :\longmapsto\:I \: \rm \: = \bigg( - \: \dfrac{cos2x}{2} + \dfrac{sin2x}{2} \bigg)_\frac{\pi}{4} ^ \frac{\pi}{3}$

$\rm :\longmapsto\:I \: =\dfrac{1}{2} \bigg( - cos\dfrac{2\pi}{3} + sin\dfrac{2\pi}{3} \bigg) - \dfrac{1}{2} \bigg( - cos\dfrac{\pi}{2} + sin\dfrac{\pi}{2}\bigg)$

$\rm :\longmapsto\:I \: =\dfrac{1}{2} \bigg(\dfrac{1}{2} + \dfrac{ \sqrt{3} }{2} \bigg) - \dfrac{1}{2} \bigg( 0 + 1\bigg)$

$\rm :\longmapsto\:I \: =\dfrac{1}{2} \bigg( \dfrac{1}{2} + \dfrac{ \sqrt{3} }{2} - 1\bigg)$

$\rm :\longmapsto\:I \: =\dfrac{1}{2} \bigg( 0.5 + \dfrac{1.732}{2} - 1\bigg)$

$\rm :\longmapsto\:I \: =\dfrac{1}{2} ( - 0.5 + 0.866)$

$\rm :\longmapsto\:I \: =\dfrac{1}{2} \times 0.366 = 0.183$

$\bf\implies \:I \: =0.2 \: (approx)$

Additional Information :-

Properties of definite integrals.

$\sf (1). \: \: \int\limits^b_a {f(x)} \, dx = \int\limits^b_a {f(t)}$

$\sf (2). \: \: \int\limits^b_a {f(x)} \, dx = -\int\limits^a_b {f(x)} </p><p>$

$\sf (3). \: \int\limits^b_a {f(x)} \, dx = \int\limits^c_a {f(x)} \, dx + \int\limits^b_c {f(x)} \, dx \ where\ \ a < c < b$

$\sf (4). \: \: \int\limits^a_0 {f(x)} \, dx =\int\limits^a_0 {f(a - x)}$

$\sf (5). \: \: \int\limits^b_a {f(x)} \, dx = \int\limits^b_a{f(a + b - x)}$