24. What is the use of the Fill handle in a Calc worksheet?
a. Copy a formula
b. Display the cell text
C. Align the text in the cell
d. Format the cell text
Answers
Answer:
We're asked to evaluate,
\displaystyle\longrightarrow I=\int\sqrt{x^2+2x+5}\ dx⟶I=∫
x
2
+2x+5
dx
Taking 5=1+4,5=1+4,
\displaystyle\longrightarrow I=\int\sqrt{x^2+2x+1+4}\ dx⟶I=∫
x
2
+2x+1+4
dx
\displaystyle\longrightarrow I=\int\sqrt{(x+1)^2+2^2}\ dx⟶I=∫
(x+1)
2
+2
2
dx
\displaystyle\longrightarrow I=\int\sqrt{2^2\left[\dfrac{(x+1)^2}{2^2}+1\right]}\ dx⟶I=∫
2
2
[
2
2
(x+1)
2
+1]
dx
\displaystyle\longrightarrow I=2\int\sqrt{\left(\dfrac{x+1}{2}\right)^2+1\right]}\ dx\quad\quad\dots(1)
Put,
\longrightarrow\dfrac{x+1}{2}=\tan\theta⟶
2
x+1
=tanθ
\longrightarrow x=2\tan\theta-1⟶x=2tanθ−1
\longrightarrow dx=2\sec^2\theta\ d\theta⟶dx=2sec
2
θ dθ
Also,
\longrightarrow\sec\theta=\sqrt{\tan^2\theta+1}⟶secθ=
tan
2
θ+1
\longrightarrow\sec\theta=\sqrt{\left(\dfrac{x+1}{2}\right)^2+1}⟶secθ=
(
2
x+1
)
2
+1
\longrightarrow\sec\theta=\dfrac{1}{2}\sqrt{x^2+2x+5}⟶secθ=
2
1
x
2
+2x+5
Then (1) becomes,
\displaystyle\longrightarrow I=4\int\sqrt{\tan^2\theta+1\right]}\ \sec^2\theta\ d\theta
\displaystyle\longrightarrow I=4\int\sec^3\theta\ d\theta\quad\quad\dots(1)⟶I=4∫sec
3
θ dθ…(1)
\displaystyle\longrightarrow I=4\int\sec\theta\ \sec^2\theta\ d\theta⟶I=4∫secθ sec
2
θ dθ
Performing integration by parts,
\displaystyle\longrightarrow I=4\left[\sec\theta\tan\theta-\int \sec\theta\tan^2\theta\ d\theta\right]⟶I=4[secθtanθ−∫secθtan
2
θ dθ]
\displaystyle\longrightarrow I=4\sec\theta\tan\theta-4\int \sec\theta(\sec^2\theta-1)\ d\theta⟶I=4secθtanθ−4∫secθ(sec
2
θ−1) dθ
\displaystyle\longrightarrow I=4\sec\theta\tan\theta-4\int(\sec^3\theta-\sec\theta)\ d\theta⟶I=4secθtanθ−4∫(sec
3
θ−secθ) dθ
\displaystyle\longrightarrow I=4\sec\theta\tan\theta-4\int\sec^3\theta\ d\theta+4\int\sec\theta\ d\theta\quad\quad\dots(2)⟶I=4secθtanθ−4∫sec
3
θ dθ+4∫secθ dθ…(2)
Adding (1) and (2) we get,
\displaystyle\longrightarrow 2I=4\sec\theta\tan\theta+4\int\sec\theta\ d\theta⟶2I=4secθtanθ+4∫secθ dθ
\displaystyle\longrightarrow I=2\sec\theta\tan\theta+2\log|\sec\theta+\tan\theta|+C'⟶I=2secθtanθ+2log∣secθ+tanθ∣+C
′
Undoing substitutions for \sec\thetasecθ and \tan\theta,tanθ,
\displaystyle\longrightarrow I=\dfrac{1}{2}(x+1)\sqrt{x^2+2x+5}+2\log\left|\dfrac{x+1+\sqrt{x^2+2x+5}}{2}\right|+C'⟶I=
2
1
(x+1)
x
2
+2x+5
+2log
∣
∣
∣
∣
∣
∣
2
x+1+
x
2
+2x+5
∣
∣
∣
∣
∣
∣
+C
′
Taking C'-2\log2=C,C
′
−2log2=C,
\displaystyle\longrightarrow\underline{\underline{I=\dfrac{1}{2}(x+1)\sqrt{x^2+2x+5}+2\log\left|x+1+\sqrt{x^2+2x+5}\right|+C}}⟶
I=
2
1
(x+1)
x
2
+2x+5
+2log
∣
∣
∣
∣
x+1+
x
2
+2x+5
∣
∣
∣