Question

If sweets are distributed among 24 children, each gets 5 sweets. How many will each get, if the number of the children is reduced by 4?
​

Answer 1

Step-by-step explanation:

Solution−

Given integral is

\rm \: \displaystyle \int_{0}^{1} \cot ^{-1}\left(1-x+x^{2}\right)d x∫

0

1

cot

−1

(1−x+x

2

)dx

We know that

\begin{gathered}\boxed{\tt{ \: {cot}^{ - 1}x = {tan}^{ - 1} \frac{1}{x} \: }} \\ \end{gathered}

cot

−1

x=tan

−1

x

1

So, using this, we get

\rm \: = \: \displaystyle \int_{0}^{1} \tan ^{-1}\bigg[\dfrac{1}{1 - x + {x}^{2} } \bigg] d x=∫

0

1

tan

−1

[

1−x+x

2

1

]dx

\rm \: = \: \displaystyle \int_{0}^{1} \tan ^{-1}\bigg[\dfrac{1 - x + x}{1 - x + {x}^{2} } \bigg] d x=∫

0

1

tan

−1

[

1−x+x

2

1−x+x

]dx

\rm \: = \: \displaystyle \int_{0}^{1} \tan ^{-1}\bigg[\dfrac{(1 - x) + x}{1 - x(1 - x)} \bigg] d x=∫

0

1

tan

−1

[

1−x(1−x)

(1−x)+x

]dx

We know

\begin{gathered}\boxed{\tt{ \: {tan}^{ - 1} \frac{x + y}{1 - xy} = {tan}^{ - 1}x + {tan}^{ - 1}y \: }} \\ \end{gathered}

tan

−1

1−xy

x+y

=tan

−1

x+tan

−1

y

Answer 2

Step-by-step explanation:

Solution−

Given integral is

\rm \: \displaystyle \int_{0}^{1} \cot ^{-1}\left(1-x+x^{2}\right)d x∫

0

1

cot

−1

(1−x+x

2

)dx

We know that

\begin{gathered}\boxed{\tt{ \: {cot}^{ - 1}x = {tan}^{ - 1} \frac{1}{x} \: }} \\ \end{gathered}

cot

−1

x=tan

−1

x

1

So, using this, we get

\rm \: = \: \displaystyle \int_{0}^{1} \tan ^{-1}\bigg[\dfrac{1}{1 - x + {x}^{2} } \bigg] d x=∫

0

1

tan

−1

[

1−x+x

2

1

]dx

\rm \: = \: \displaystyle \int_{0}^{1} \tan ^{-1}\bigg[\dfrac{1 - x + x}{1 - x + {x}^{2} } \bigg] d x=∫

0

1

tan

−1

[

1−x+x

2

1−x+x

]dx

\rm \: = \: \displaystyle \int_{0}^{1} \tan ^{-1}\bigg[\dfrac{(1 - x) + x}{1 - x(1 - x)} \bigg] d x=∫

0

1

tan

−1

[

1−x(1−x)

(1−x)+x

]dx

We know

\begin{gathered}\boxed{\tt{ \: {tan}^{ - 1} \frac{x + y}{1 - xy} = {tan}^{ - 1}x + {tan}^{ - 1}y \: }} \\ \end{gathered}

tan

−1

1−xy

x+y

=tan

−1

x+tan

−1

y