3 years ago

What is the answer to this question

Mathematics

1 answer:

iren2701 [21]3 years ago

7 0

Answer:

Step-by-step explanation:

The integral decomposes into the sum ...

$\displaystyle\int_{-2}^2{\left(x^3\cos{\dfrac{x}{2}}\right)\sqrt{4-x^2}}\, dx+\dfrac{1}{2}\int_{-2}^2{\sqrt{4-x^2}}\, dx$

The first term of this sum is the integral of an odd function, so is zero. The second term of this sum is 1/2 the area under a semicircle of radius 2, so is ...

A = (1/2)(1/2)π·2² = π

The value of the integral is π.

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Simplify sin^2y/sec^2 y−1 to a single trigonometric function

liubo4ka [24]

Answer:

$\frac{ { \sin}^{2} y}{ { \sec}^{2}y - 1 } = { \cos }^{2} y$

Step-by-step explanation:

We know that ${ \tan }^{2} y = { \sec }^{2} y - 1$

Also , ${ \tan}^{2} y = \frac{ { \sin }^{2} y}{ { \cos }^{2}y }$

So ,

$\frac{ { \sin }^{2}y }{ { \sec }^{2}y - 1 } = \frac{ { \sin}^{2} y}{ { \tan }^{2} y} = \frac{ { \sin }^{2}y }{ \frac{ { \sin}^{2} y}{ { \cos}^{2}y } } = { \cos }^{2} y$