Example.1
If
I(n) =
, n ∈ N, n > 3,
then find the value of I(n)-
.
Solution
I(n) =
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image004.gif)
= I(n – 2) – ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image005.gif)
= I(n – 2) – q.cosq.![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image006.gif)
= I(n – 2) – ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image007.gif)
= I(n – 2) – ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image008.gif)
&⇒ ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image009.gif)
&⇒ I(n) – I(n –2) .
.
Example.2
If a > 1, then evaluate
.
Solution
Let
I (a) =
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image013.gif)
I¢(a) =
, Let x = sin q, dx = cos q dq
&⇒
I¢(a) =![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image015.gif)
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image017.gif)
\
I (a) = p sin–1(a)
+ c where for a = 0, I(a) = 0
&⇒
c = 0
\
I (a) = p sin–1
(a)
Example.3
Evaluate
.
Solution
Let I = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image019.gif)
Also I = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image020.gif)
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image021.gif)
Adding, we get 2I =
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image022.gif)
=
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image024.gif)
=
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image026.gif)
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image027.gif)
&⇒ I = ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image028.gif)
Example.4
Find the mistake in
the evaluation of the following integral and correct it. .
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image029.gif)
Solution
The function y =
is continuous in the interval (0, p). Also the curve, from x = 0 to p, lies above the x-axis.
Hence area under the
curve cannot be zero. We, therefore, use the property of definite integral and
write .
![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image031.gif)
= 2
=
=
.
Example.5
Prove that for any
positive integer k ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image035.gif)
Hence prove that
.
Solution
To prove ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image037.gif)
Let t1 = 2 sinx cosx = sin2x
t2 = 2 sinx cos3x = sin4x – sin2x
t3 = 2 sinx cos5x = sin6x – sin4x
…………………………………..
…………………………………..
tk = 2
sinx cos(2k-1)x = sin2kx – sin(2k –2)x
Adding, we get
R.H.S = Sk = sin2kx .
Now,
= ![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image039.gif)
=
dx
=
+
+...+![](http://www.quizsolver.com/radix/dth/notif/Definite%20SUB%202_files/image043.gif)
But, we know that
=
0 " n ∈ I and n > 0.
Hence,
+
0 =
.