also record your er. Record observation regarding of melon or gourd flower as observations in the table given below. (You may flowers already collected by you earlier in this chapter). S.No. Name of the No. of Sepalss No. of Petals No. of Stamens No. of Pistii Flower (Calyx) (Corolla) (Andioecium)||(Gynoecium) 2) 5 14 Reproduction in Plan
Answers
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
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