If -sin 101 x sin 99 x d x-sin p x -sin x9-r-c- thenA- r-100B- p-101C- q-99D- p-q-2 r

The correct options are A r = 100 D p + q = 2rI = ∫ sin(100x + x).(sin x)^99dx = sin(100x). (sin x)^99 cos x dx + ∫ cos (100x).(sin x)^100 dx = sin(100 x).(si ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Limits

If ∫sin 101 x...

Question

If $\int$ sin (101 x) $s i n^{99}$ x dx = $\frac{s i n (p x) . (s i n x)^{q}}{r} + c,$ then

r = 100

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p = 101

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q = 99

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p + q = 2r

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Solution

The correct options are
A r = 100
D p + q = 2r
$I = \int s i n (100 x + x) . (s i n x)^{99} d x = s i n (100 x) . (s i n x)^{99} c o s x d x + \int c o s (100 x) . (s i n x)^{100} d x = \frac{s i n (100 x) . (s i n x)^{100}}{100} + c$
[Applying by parts]

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If ∫ sin (101 x) sin99 x dx = sin(px).(sin x)qr+c, then

The correct options are A r = 100 D p + q = 2rI=∫sin(100x+x).(sin x)99dx= sin(100x).(sin x)99cos x dx+∫cos(100x).(sin x)100dx=sin(100 x).(sin x)100100+c [Applying by parts]

If $\int$ sin (101 x) $s i n^{99}$ x dx = $\frac{s i n (p x) . (s i n x)^{q}}{r} + c,$ then

The correct options are
A r = 100
D p + q = 2r
$I = \int s i n (100 x + x) . (s i n x)^{99} d x = s i n (100 x) . (s i n x)^{99} c o s x d x + \int c o s (100 x) . (s i n x)^{100} d x = \frac{s i n (100 x) . (s i n x)^{100}}{100} + c$
[Applying by parts]