Question

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Answer 1

Ответ:

1.

$\int\limits ^{ \frac{\pi}{3} } _ { 0} \frac{dx}{ { \cos }^{2} (x)} = tg(x) | ^{ \frac{\pi}{3} } _ {0} = \\ = tg( \frac{\pi}{3} ) - tg(0) = \sqrt{3} - 0 = \sqrt{3}$

2.

$\int\limits ^{ \frac{\pi}{2} } _ {0 } { \cos }^{2}(x)dx =\int\limits ^{ \frac{\pi}{2} } _ {0 } \frac{1 + \cos(2x) }{2} dx = \\ = \frac{1}{2}(\int\limits ^{ \frac{\pi}{2} } _ {0 }dx + \frac{1}{2} \int\limits ^{ \frac{\pi}{2} } _ {0 } \cos(2x)d(2x)) = \\ = ( \frac{1}{2} x + \frac{1}{4} \sin(2x)) | ^{ \frac{\pi}{2} } _ {0} = \\ = \frac{\pi}{4} + \frac{1}{4} \sin(\pi) - 0 - \frac{1}{4} \sin(0) = \\ = \frac{\pi}{4} + 0 - 0= \frac{\pi}{4}$

3.

$\int\limits ^{ \frac{\pi}{3} } _ { \frac{\pi}{4} } \frac{dx}{ { \sin }^{2} (x)} = - ctg(x)| ^{ \frac{\pi}{3} } _ { \frac{\pi}{4} } = \\ = - ctg( \frac{\pi}{3} ) + ctg( \frac{\pi}{4} ) = - \frac{ \sqrt{3} }{3} + 1 = \\ = \frac{3 - \sqrt{3} }{3}$

4.

$\int\limits ^{1} _ {0 } { \sin }^{2} (x)dx = \int\limits ^{1} _ {0 } \frac{1 - \cos(2x) }{2} dx = \\ = \frac{1}{2} (\int\limits ^{1} _ {0 }dx - \frac{1}{2} \int\limits ^{1} _ {0 } \cos(2x) d(2x)) = \\ = (\frac{1}{2} x - \frac{1}{4} \sin(2x)) | ^{ 1} _ { 0} = \\ = \frac{1}{2} - \frac{1}{4} \sin(2) - 0 + \frac{1}{4} \sin(0) = \\ = \frac{1}{2} - \frac{1}{4} \sin(2)$